Trigonometric Ratios Upto Transformations (VSAQs)

Maths-1A | 6. Trigonometric Ratios Upto Transformations – VSAQs:
Welcome to VSAQs in Chapter 6: Trigonometric Ratios Upto Transformations. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

Overview:

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: $$\text{Period} = \frac{2\pi}{3}$$

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

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

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 $$\frac{2\pi}{4} = \frac{\pi}{2}$$

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

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

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 $$|5| = 5$$

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

$$\pi/5$$

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:

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

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.

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, $$\sin(3\pi x)$$

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.

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.

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.

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.

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:

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

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

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

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

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

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

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

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

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

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

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: $$\tan(90^\circ – 36^\circ) = \cot36^\circ$$

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^∘: $$\tan \theta = \tan(56^\circ) = \tan(180^\circ + 56^\circ) = \tan 236^\circ$$

$$\theta = 236^\circ$$

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: $$\sin (-\theta) = -\sin (\theta)$$ $$= -\sin 10^\circ + \sin 10^\circ$$

$$= 0$$

$$\text{RHS} = 0$$

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^∘ $$= 2\cdot\frac{1}{2}\cdot\cos18^\circ – \cos18^\circ$$

$$= \cos18^\circ – \cos18^\circ$$

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

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: $$\tan 50^\circ – \tan 40^\circ / 1 + \tan 50^\circ \tan 40^\circ = \tan 10^\circ$$

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

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

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: $$\tan 70^\circ – \tan 20^\circ / 1 + \tan 70^\circ \tan 20^\circ = \tan 50^\circ$$

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

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

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: $$= \frac{-\lambda + \frac{1}{\lambda}}{1 + 1}$$ $$= \frac{-\lambda^2 + 1}{\lambda(2)}$$ $$= \frac{1 – \lambda^2}{2\lambda}$$

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

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

$$GE = \cos(48°) \cdot \cos(12°)$$

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

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

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

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

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

VSAQ-21 : Prove that sin²42°- cos²78° = (√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 $$\sin^2 A – \sin^2 B = \sin(A + B)\sin(A – B)$$

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

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

VSAQ-22 : Find the value of cos²52 1°/2 – sin²22 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: $$A + B = 52\frac{1}{2}^\circ + 22\frac{1}{2}^\circ = 75^\circ$$ $$A – B = 52\frac{1}{2}^\circ – 22\frac{1}{2}^\circ = 30^\circ$$

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

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

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

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

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

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

$$x^2 = a^2 + b^2 – c^2$$

$$x = \pm \sqrt{a^2 + b^2 – c^2}$$

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

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

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

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

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

$$x^2 + 25 = 25$$

$$x^2 = 0$$

$$x = 0$$

$$4\sin\theta – 3\cos\theta = 0$$