24 Most VSAQ’s of Trigonometric Ratios Upto Transformations Chapter in Inter 1st Year Maths-1A (TS/AP)

2 Marks

VSAQ-1 : Find the period of f(x) = cos (3x + 5) + 7

$$f(x) = \cos(3x + 5) + 7$$

Step 1: Identify k from the given function. In $$f(x) = \cos(3x + 5) + 7$$ $$k = 3$$

Step 2: Apply the formula for the period: $$\text{Period} = \frac{2\pi}{|k|}$$

Step 3: Substitute k=3 into the formula: $$\text{Period} = \frac{2\pi}{|3|}$$

Step 4: Calculate the period: $$\text{Period} = \frac{2\pi}{3}$$

$$f(x) = \cos(3x + 5) + 7$$

$$\frac{2\pi}{3}$$

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VSAQ-2 : Find the period of f(x) = cos (4x+9)/5)

$$f(x) = \cos(4x + \frac{9}{5})$$

Step 1 : Identify k : In $$f(x) = \cos(4x + \frac{9}{5})$$ $$k = 4$$

Step 2 : Apply the formula: Period of $$\cos(kx + l)$$ = $$\frac{2\pi}{|k|}$$

Step 3 : Substitute k : Period $$\frac{2\pi}{|4|}$$

Step 4 : Calculate: Period $$\frac{2\pi}{4} = \frac{\pi}{2}$$

$$f(x) = \cos(4x + \frac{9}{5})$$

$$\frac{\pi}{2}$$

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VSAQ-3 : Find the period of tan 5x

$$f(x) = \tan(5x)$$

Step 1 : Identify the Function: The given function is $$f(x) = \tan(5x)$$

Step 2 : Identify the Coefficient of x: In this function, the coefficient of x inside the tangent function is 5.

Step 3 : Use the Period Formula: The period of a tangent function, $$\tan(kx)$$ is given by $$\pi/|k|$$

where k is the coefficient of x.

Step 4 : Substitute k with 5: Replace k in the formula with 5, giving $$\pi/|5|$$

Step 5 : Calculate the Period: Since $$|5| = 5$$

$$f(x) = \tan(5x)$$

$$\pi/5$$

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VSAQ-4 : Find the period of tan ( x + 4x + 9x + … + n²x)

$$\tan(x + 4x + 9x + \ldots + n^2x)$$

Step 1 : Sum the Series: Sum the series inside the tangent: $$1^2 + 2^2 + 3^2 + \ldots + n^2$$

Step 2 : Apply Formula: Use the sum formula for squares: $$\frac{n(n + 1)(2n + 1)}{6}$$

Step 3 : Rewrite Function: The function becomes $$\tan\left(\frac{n(n + 1)(2n + 1)}{6}x\right)$$

Step 4 : Determine Coefficient: The coefficient of x is $$\frac{n(n + 1)(2n + 1)}{6}$$

Step 5 : Calculate Period: Use $$\pi/|k|$$ with $$k = \frac{n(n + 1)(2n + 1)}{6}$$ to find the period:

$$\frac{6\pi}{n(n + 1)(2n + 1)}$$

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VSAQ-5 : Find a cosine function whose period is 7

To find the required cosine function with a period of 7

Step 1 : Identify the Function: We’re looking for a cosine function of the form $$\cos(kx)$$

Step 2 : Use the Period Formula for Cosine: The period of $$\cos(kx)$$ is given by $$2\pi/k$$

Step 3 : Set the Period Equal to 7: To find the specific function, set the period formula equal to 7: $$2\pi/k = 7$$

Step 4 : Solve for k: Rearrange the equation to solve for k: $$k = 2\pi/7$$

Step 5 : Identify the Required Cosine Function: Substitute k back into the cosine function to get $$\cos(2\pi/7)x$$

Thus, the required cosine function is $$\cos(2\pi/7)x$$ with a period of 7.

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VSAQ-6 : Find a sine function whose period is 2/3

Let sinkx be the required sine function

Step 1 : Identify the Function Type: We are looking for a sine function, $$\sin(kx)$$

Step 2 : Period Formula for Sine Function: The period of $$\sin(kx)$$ is $$2\pi/k$$

Step 3 : Set the Given Period: We’re given the period as $$2/3$$ so we set $$2\pi/k = 2/3$$

Step 4 : Solve for k: To find k, rearrange the equation: $$k = 2\pi \times (3/2)$$

$$k = 3\pi$$

Thus, the required sine function, $$\sin(3\pi x)$$

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VSAQ-7 : Find the maximum and minimum value of f(x) = 3cosx + 4sinx

Given function is $$3\cos(x) + 4\sin(x)$$

Step 1 : Identify the Function: The function is $$3\cos(x) + 4\sin(x)$$

Step 2 : Compare to Standard Form: Compare it to the standard form $$a\cos(x) + b\sin(x) + c$$ giving $$a=3$$ $$b=4$$ $$c=0$$

Step 3 : Calculate $$\sqrt{a^2 + b^2}$$ $$\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Step 4 : Find Maximum Value: The maximum value of the function is given by $$c + \sqrt{a^2 + b^2}$$ Substituting the values, we get $$0 + 5 = 5$$

Step 5 : Find Minimum Value: The minimum value of the function is given by $$c – \sqrt{a^2 + b^2}$$ Substituting the values, we get $$0 – 5 = -5$$

Thus, the maximum value of the function $$3\cos(x) + 4\sin(x)$$ is 5, and the minimum value is -5.

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VSAQ-8 : Find the maximum and minimum values of f(x) = 3sinx – 4cosx

Given function is $$3\sin(x) – 4\cos(x)$$

Step 1 : Identify the Function: The function is $$3\sin(x) – 4\cos(x)$$

Step 2 : Compare to Standard Form: When comparing to the standard form $$a\sin(x) + b\cos(x) + c$$ we identify $$a=3$$ $$b=-4$$ $$c=0$$

Step 3 : Calculate $$\sqrt{a^2 + b^2}$$ $$\sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Step 4 : Find Maximum Value: The maximum value is given by $$c + \sqrt{a^2 + b^2}$$ Substituting the known values, $$0 + 5 = 5$$

Step 5 : Find Minimum Value: The minimum value is given by $$c – \sqrt{a^2 + b^2}$$ Substituting the known values, $$0 – 5 = -5$$

Thus, the maximum value of the function $$3\sin(x) – 4\cos(x)$$ is 5, and the minimum value is -5.

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VSAQ-9 : Find the maximum and minimum value of f(x) = 5sinx + 12cosx – 13

Given function $$f(x) = 5\sin(x) + 12\cos(x) – 13$$

Step 1 : Identify the Function: The function is $$f(x) = 5\sin(x) + 12\cos(x) – 13$$

Step 2 : Compare to Standard Form: When comparing to the standard form $$a\sin(x) + b\cos(x) + c$$ we identify $$a=5$$ $$b=12$$ $$c=-13$$

Step 3 : Calculate $$\sqrt{a^2 + b^2}$$ $$\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$

Step 4 : Find Maximum Value: The maximum value is given by $$c + \sqrt{a^2 + b^2}$$ Substituting the known values, $$-13 + 13 = 0$$

Step 5 : Find Minimum Value: The minimum value is given by $$c – \sqrt{a^2 + b^2}$$ Substituting the known values, $$-13 – 13 = -26$$

Thus, the maximum value of the function $$5\sin(x) + 12\cos(x) – 13$$ is 0, and the minimum value is -26.

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VSAQ-10 : Find the range of 7cosx – 24sinx + 5

Given function is $$f(x) = 7\cos(x) – 24\sin(x) + 5$$

Step 1 : Identify the Function: The function is $$7\cos(x) – 24\sin(x) + 5$$

Step 2 : Compare to Standard Form: The standard form is $$a\sin(x) + b\cos(x) + c$$ which gives us $$a=-24$$ $$b=7$$ $$c=5$$

Step 3 : Calculate $$\sqrt{a^2 + b^2}$$ $$\sqrt{(-24)^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25$$

Step 4 : Find Maximum Value: The maximum value of the function is $$c + \sqrt{a^2 + b^2}$$ Substituting the known values, we get $$5 + 25 = 30$$

Step 5 : Find Minimum Value: The minimum value of the function is $$c – \sqrt{a^2 + b^2}$$ Substituting the known values, we get $$5 – 25 = -20$$

Therefore, the range of the function $$7\cos(x) – 24\sin(x) + 5$$ is from -20 to 30.

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VSAQ-11 : Find sin330°.cos120° + cos210°.sin300°

$$\sin(330^\circ) = \sin(360^\circ – 30^\circ) = -\sin(30^\circ) = -\frac{1}{2}$$

$$\cos(120^\circ) = \cos(180^\circ – 60^\circ) = -\cos(60^\circ) = -\frac{1}{2}$$

$$\cos(210^\circ) = \cos(180^\circ + 30^\circ) = -\cos(30^\circ) = -\frac{\sqrt{3}}{2}$$

$$\sin(300^\circ) = \sin(360^\circ – 60^\circ) = -\sin(60^\circ) = -\frac{\sqrt{3}}{2}$$

So the given expression (G.E) can be calculated as:

$$\text{G.E} = \left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) + \left(-\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right)$$

$$\text{G.E} = \frac{1}{4} + \frac{3}{4}$$

$$\text{G.E} = \frac{4}{4}$$

$$\text{G.E} = 1$$

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VSAQ-12 : S.T cos340° cos40° + sin200° sin140°= 1/2

$$\text{L.H.S} = \cos(340^\circ) \cdot \cos(40^\circ) + \sin(200^\circ) \cdot \sin(140^\circ)$$

$$= \cos(360^\circ – 20^\circ) \cdot \cos(40^\circ) + \sin(180^\circ + 20^\circ) \cdot \sin(180^\circ – 40^\circ)$$

$$= \cos(20^\circ) \cdot \cos(40^\circ) – \sin(20^\circ) \cdot \sin(40^\circ)$$

$$= \cos(20^\circ + 40^\circ) = \cos(60^\circ)$$

$$\text{R.H.S} = \frac{1}{2}$$

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VSAQ-13 : Prove that (cos9°+sin9°)/(cos9°-sin9°)=cot36°

Step 1 : Starting with the given expression: $$\text{L.H.S} = \frac{\cos9^\circ + \sin9^\circ}{\cos9^\circ – \sin9^\circ}$$

Step 2 : The expression is then manipulated as follows: $$\frac{\cos9^\circ}{\cos9^\circ} + \frac{\sin9^\circ}{\cos9^\circ} \div \frac{\cos9^\circ}{\cos9^\circ} – \frac{\sin9^\circ}{\cos9^\circ} = 1 + \tan9^\circ \div 1 – \tan9^\circ$$

Step 3 : Further simplification using trigonometric identities: $$\tan45^\circ + \tan9^\circ \div 1 – \tan45^\circ\tan9^\circ = \tan(45^\circ+9^\circ) = \tan54^\circ$$

Step 4 : Final steps leading to the conclusion: $$\tan(90^\circ – 36^\circ) = \cot36^\circ$$

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VSAQ-14 : If tanθ = (cos11°+sin11°)/(cos11°-sin11°) and θ in the third quadrant find θ

Given: $$\tan \theta = \frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ – \sin 11^\circ}$$

Step 1 : Starting Point: $$\tan \theta = \frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ – \sin 11^\circ}$$

Step 2 : Divide Numerator and Denominator by cos11^∘: $$\tan \theta = \frac{1 + \frac{\sin 11^\circ}{\cos 11^\circ}}{1 – \frac{\sin 11^\circ}{\cos 11^\circ}}$$

Step 3 : Simplify Using tan11^∘: $$\tan \theta = \frac{1 + \tan 11^\circ}{1 – \tan 11^\circ}$$

Step 4 : Recognize as a Known Trigonometric Identity: $$\tan \theta = \tan(45^\circ + 11^\circ)$$

Step 5 : Simplify: $$\tan \theta = \tan 56^\circ$$

Step 6 : Periodicity of Tan Function: The tangent function has a period of 180^∘, meaning $$\tan(\alpha) = \tan(\alpha + 180^\circ)$$

Step 7 : Extend to 236^∘: $$\tan \theta = \tan(56^\circ) = \tan(180^\circ + 56^\circ) = \tan 236^\circ$$

$$\theta = 236^\circ$$

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VSAQ-15 : Prove that sin50° – sin70° + sin10° = 0

Given: $$\text{LHS} = (\sin 50^\circ – \sin 70^\circ) + \sin 10^\circ$$

We’ll use the trigonometric identity: $$\sin A – \sin B = 2 \cos \left(\frac{A + B}{2}\right) \sin \left(\frac{A – B}{2}\right)$$

Step 1 : Starting Point: $$(\sin 50^\circ – \sin 70^\circ) + \sin 10^\circ$$

Step 2 : Apply the Identity: = $$2\cos\left(\frac{50^\circ + 70^\circ}{2}\right)\sin\left(\frac{50^\circ – 70^\circ}{2}\right) + \sin 10^\circ$$

Step 3 : Simplify Inside the Cosine and Sine: = $$2\cos\left(\frac{120^\circ}{2}\right)\sin\left(\frac{-20^\circ}{2}\right) + \sin 10^\circ$$ $$= 2\cos 60^\circ \sin (-10^\circ) + \sin 10^\circ$$

Step 4 : Apply Exact Values for Cosine: $$= 2 \times \frac{1}{2} \sin (-10^\circ) + \sin 10^\circ$$ $$= \sin (-10^\circ) + \sin 10^\circ$$

Step 5 : Using the Odd Function Property of Sine: $$\sin (-\theta) = -\sin (\theta)$$ $$= -\sin 10^\circ + \sin 10^\circ$$

$$= 0$$

$$\text{RHS} = 0$$

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VSAQ-16 : Show that cos42° + cos78° + cos162° = 0

Given: $$\text{L.H.S} = \cos 42^\circ + \cos 78^\circ + \cos 162^\circ$$

Step 1: Rewrite each term using known angles $$= \cos(60^\circ – 18^\circ) + \cos(60^\circ + 18^\circ) + \cos(180^\circ – 18^\circ)$$

Step 2: Apply sum-to-product identities $$\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A – B}{2} \right)$$

$$= 2\cos60^\circ\cos18^\circ + \cos(180^\circ – 18^\circ)$$

Step 3: Simplify the third term $$\cos(180^\circ – 18^\circ) = -\cos18^\circ$$ $$\cos(180^\circ – \theta) = -\cos \theta$$

Step 4: Simplify the expression using the value of cos⁡60^∘ $$= 2\cdot\frac{1}{2}\cdot\cos18^\circ – \cos18^\circ$$

$$= \cos18^\circ – \cos18^\circ$$

$$= 0 = \text{R.H.S}$$

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VSAQ-17 : Prove that tan50° – tan 40° = 2tan10°

Given that we start with the difference of two angles: $$50^\circ – 40^\circ = 10^\circ$$

Step 1 : This leads us to consider the tangent of this difference: $$\tan(50^\circ – 40^\circ) = \tan 10^\circ$$

Step 2 : From the tangent difference identity, we know that: $$\tan(A – B) = \frac{\tan A – \tan B}{1 + \tan A \tan B}$$

Step 3 : Applying this to our given angles: $$\tan 50^\circ – \tan 40^\circ / 1 + \tan 50^\circ \tan 40^\circ = \tan 10^\circ$$

$$\frac{\tan 50^\circ – \tan 40^\circ}{1 + \tan 50^\circ \tan 40^\circ} = \tan 10^\circ$$

$$\tan 50^\circ – \tan 40^\circ = 2 \tan 10^\circ$$

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VSAQ-18 : Prove that tan70° – tan20° = 2tan50°

Given that we start with the difference between two angles: $$70^\circ – 20^\circ = 50^\circ$$

Step 1 : This leads us to the tangent of this difference: $$\tan(70^\circ – 20^\circ) = \tan 50^\circ$$

Step 2 : By the tangent difference formula, we have: $$\tan(A – B) = \frac{\tan A – \tan B}{1 + \tan A \tan B}$$

Step 3 : Applying this formula with A=70^∘ and B=20^∘, we get: $$\tan 70^\circ – \tan 20^\circ / 1 + \tan 70^\circ \tan 20^\circ = \tan 50^\circ$$

$$\frac{\tan 70^\circ – \tan 20^\circ}{1 + \tan 70^\circ \tan 20^\circ} = \tan 50^\circ$$

$$\tan 70^\circ – \tan 20^\circ = 2 \tan 50^\circ$$

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VSAQ-19 : If tan20° = λ then show that (tan160°-tan110°)/(1+tan160°.tan110°)=(1-λ²)/2

Given:

$$\tan 160^\circ = \tan(180^\circ – 20^\circ) = -\tan 20^\circ = -\lambda$$

$$\tan 110^\circ = \tan(90^\circ + 20^\circ) = \cot 20^\circ = \frac{1}{\tan 20^\circ} = -\frac{1}{\lambda}$$

Step 1 : Now, let’s address the expression: $$\text{L.H.S} = \frac{\tan 160^\circ – \tan 110^\circ}{1 + \tan 160^\circ \tan 110^\circ}$$

Step 2 : Substituting the given values: = $$\frac{-\lambda – \left(-\frac{1}{\lambda}\right)}{1 + (-\lambda)\left(-\frac{1}{\lambda}\right)}$$

Step 3 : Simplify the expression: $$= \frac{-\lambda + \frac{1}{\lambda}}{1 + 1}$$ $$= \frac{-\lambda^2 + 1}{\lambda(2)}$$ $$= \frac{1 – \lambda^2}{2\lambda}$$

$$\text{R.H.S} = \frac{1 – \lambda^2}{2\lambda}$$

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VSAQ-20 : Find the value of Cos 48° . Cos 12°

$$GE = \cos(48°) \cdot \cos(12°)$$

$$= \frac{1}{2} [2\cos(48°) \cos(12°)]$$

$$= \frac{1}{2} [\cos(60°) + \cos(36°)]$$

$$= \frac{1}{2} [\frac{1}{2} + \sqrt{5} + \frac{1}{4}]$$

$$= \frac{1}{2} [2 + \sqrt{5} + \frac{1}{4}]$$

$$= \frac{3 + \sqrt{5}}{8}$$

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VSAQ-21 : Prove that sin² 42°- cos² 78° = (√5+1)/8

Step 1 : Starting Expression: $$\text{L.H.S} = \sin^2 42^\circ – \cos^2 78^\circ$$

Step 2 : Using the Complementary Angle Identity: $$\cos^2 78^\circ = \cos^2(90^\circ – 12^\circ) = \sin^2 12^\circ$$ $$\text{L.H.S} = \sin^2 42^\circ – \sin^2 12^\circ$$

Step 3 : Applying the Identity $$\sin^2 A – \sin^2 B = \sin(A + B)\sin(A – B)$$

$$\text{L.H.S} = \sin(42^\circ + 12^\circ)\sin(42^\circ – 12^\circ) = \sin54^\circ\sin30^\circ$$

$$(\sqrt{5} + 1)/4$$ $$\sqrt{5} + 1/8$$

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VSAQ-22 : Find the value of cos² 52 1°/2 – sin² 22 1°/2

Step 1 : Given expression: $$\cos^2(52\frac{1}{2}^\circ) – \sin^2(22\frac{1}{2}^\circ)$$

Step 2 : Apply the identity: $$\cos^2 A – \sin^2 B = \cos(A + B)\cos(A – B)$$

Here, $$A = 52\frac{1}{2}^\circ$$ $$B = 22\frac{1}{2}^\circ$$ so we calculate $$A + B$$ and $$A – B$$

Step 3 : Calculate A+B and A−B: $$A + B = 52\frac{1}{2}^\circ + 22\frac{1}{2}^\circ = 75^\circ$$ $$A – B = 52\frac{1}{2}^\circ – 22\frac{1}{2}^\circ = 30^\circ$$

$$(\sqrt{3} – 1) / (2\sqrt{2}) \times \sqrt{3}/2 = (3 – \sqrt{3}) / (4\sqrt{2})$$

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VSAQ-23 : If acosθ-bsinθ=c, then show that asinθ + bcosθ = ± √(a² + b² – c²)

Let $$asin\theta + bcos\theta = x$$

Given that $$acos\theta – bsin\theta = c$$

Squaring and adding, we get

$$x^2 + c^2 = (asin\theta + bcos\theta)^2 + (acos\theta – bsin\theta)^2$$

$$= x^2 + c^2 = (a^2sin^2\theta + 2absin\theta cos\theta + b^2cos^2\theta) + (a^2cos^2\theta – 2absin\theta cos\theta + b^2sin^2\theta)$$

$$= a^2(sin^2\theta + cos^2\theta) + b^2(sin^2\theta + cos^2\theta) = a^2 + b^2$$

$$x^2 + c^2 = a^2 + b^2$$

$$x^2 = a^2 + b^2 – c^2$$

$$x = \pm \sqrt{a^2 + b^2 – c^2}$$

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VSAQ-24 : If 3sinθ+4cosθ=5, then find the value of 4sinθ-3cosθ

Let $$4\sin\theta – 3\cos\theta = x$$

Given that $$3\sin\theta + 4\cos\theta = 5$$

Squaring and adding we get

$$x^2 + 5^2 = (4\sin\theta – 3\cos\theta)^2 + (3\sin\theta + 4\cos\theta)^2$$

$$= x^2 + 25 = (16\sin^2\theta – 24\sin\theta\cos\theta + 9\cos^2\theta) + (9\sin^2\theta + 24\sin\theta\cos\theta + 16\cos^2\theta)$$

$$= 25\sin^2\theta + 25\cos^2\theta$$

$$= 25(\sin^2\theta + \cos^2\theta) = 25(1) = 25$$

$$x^2 + 25 = 25$$

$$x^2 = 0$$

$$x = 0$$

$$4\sin\theta – 3\cos\theta = 0$$