Trigonometric Functions MCQ Quiz - Objective Question with Answer for Trigonometric Functions - Download Free PDF

Concept:

• Range of Sine Function: The output of the sine function is always between −1 and 1, i.e., sin(θ) ∈ [−1, 1] for all real θ.
• Quadratic Function: A quadratic function of the form ax² + bx + c represents a parabola. If a > 0, the parabola opens upward, and its minimum value occurs at x = −b / 2a.
• Key Idea: To find how many solutions exist for sin(expression) = quadratic, we determine how many values of x make the quadratic expression lie within [−1, 1].

Calculation:

Given,

\(\sin (\frac{\pi x }{3\sqrt{2}}) = x^2-4x+6 \)   

Let f(x) = x² − 4x + 6

Minimum of f(x) occurs at:

x = 4 / 2 = 2

⇒ f(2) = (2)² − 4×2 + 6 = 4 − 8 + 6 = 2

Since the parabola opens upward, the range of f(x) is [2, ∞)

But, sin(θ) ∈ [−1, 1]

⇒ The equation has solutions only if x² − 4x + 6 ∈ [−1, 1]

But f(x) ≥ 2 for all x, and 2 > 1

⇒ No value of x satisfies f(x) ∈ [−1, 1]

∴ Number of real solutions is zero.

Calculation:

Given,

\(x = \sec\theta - \cos\theta\)

\(y = \sec^4\theta - \cos^4\theta\)

We know from before:

\(\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = \frac{16\,(y^{2}+4)}{\,x^{2}+4\,} \)

Differentiate both sides w.r.t. x:

\(\displaystyle \frac{d}{dx}\Bigl[(x^{2}+4)\bigl(\tfrac{dy}{dx}\bigr)^{2}\Bigr] = \frac{d}{dx}\bigl[16\,(y^{2}+4)\bigr] \)

This yields the differential relation:

\(\displaystyle (x^{2}+4)\,\frac{d^{2}y}{dx^{2}} \;+\;x\,\frac{dy}{dx} \;-\;16\,y \;=\;0\)

The required expression is

\(\displaystyle \frac{x^{2}+4}{y^{2}+4}\,\frac{dy}{dx} \Bigl[(x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y\Bigr] \)

From the differential relation,

\(\displaystyle (x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y = -\,x\,\frac{dy}{dx} \)

So the expression becomes

\(\displaystyle \frac{x^{2}+4}{y^{2}+4}\,\frac{dy}{dx}\, \bigl(-x\,\tfrac{dy}{dx}\bigr) = -\,x\,\frac{x^{2}+4}{y^{2}+4} \Bigl(\frac{dy}{dx}\Bigr)^{2} \)

Using \(\bigl(\tfrac{dy}{dx}\bigr)^{2} = \tfrac{16\,(y^{2}+4)}{x^{2}+4}\), all factors cancel except -16x.

∴  The value of the given expression is  \(-16x\).

Hence, the correct answer is Option 3. 

Calculation:

Given,

\(x = \sec\theta - \cos\theta \)

\(y = \sec^4\theta - \cos^4\theta\)

Compute derivatives w.r.t. θ" id="MathJax-Element-5-Frame" role="presentation" style="position: relative;" tabindex="0">θθ $\theta$ :

\(\dfrac{dx}{d\theta} = \sec\theta\,\tan\theta + \sin\theta\)

\(\dfrac{dy}{d\theta} = 4\sec^4\theta\,\tan\theta \;+\; 4\cos^3\theta\,\sin\theta\)

From the ratio \(\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)

\(\dfrac{dy}{dx} = \frac{4\bigl(\sec^4\theta\,\tan\theta + \cos^3\theta\,\sin\theta\bigr)} {\sec\theta\,\tan\theta + \sin\theta}\)

Using identities \(x^2 = \sec^2\theta + \cos^2\theta - 2\), \(\sec^3\theta + \cos^5\theta)^2 = y^2 + 4\), and \((1 + \cos^2\theta)^2 = x^2 + 4,\)

\(\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = 16 \,\frac{\,y^{2} + 4\,}{\,x^{2} + 4\,} \)

∴  \(\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = 16\,\frac{y^{2} + 4}{x^{2} + 4} \)

Hence, the correct answer is Option 3.

Calculation:

Given,

\(x = \sec\theta - \cos\theta\)

\(y = \sec^4\theta - \cos^4\theta\)

We know from before:

\(\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = \frac{16\,(y^{2}+4)}{\,x^{2}+4\,} \)

Differentiate both sides w.r.t. x:

\(\displaystyle \frac{d}{dx}\Bigl[(x^{2}+4)\bigl(\tfrac{dy}{dx}\bigr)^{2}\Bigr] = \frac{d}{dx}\bigl[16\,(y^{2}+4)\bigr] \)

This yields the differential relation:

\(\displaystyle (x^{2}+4)\,\frac{d^{2}y}{dx^{2}} \;+\;x\,\frac{dy}{dx} \;-\;16\,y \;=\;0\)

The required expression is

\(\displaystyle \frac{x^{2}+4}{y^{2}+4}\,\frac{dy}{dx} \Bigl[(x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y\Bigr] \)

From the differential relation,

\(\displaystyle (x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y = -\,x\,\frac{dy}{dx} \)

So the expression becomes

\(\displaystyle \frac{x^{2}+4}{y^{2}+4}\,\frac{dy}{dx}\, \bigl(-x\,\tfrac{dy}{dx}\bigr) = -\,x\,\frac{x^{2}+4}{y^{2}+4} \Bigl(\frac{dy}{dx}\Bigr)^{2} \)

Using \(\bigl(\tfrac{dy}{dx}\bigr)^{2} = \tfrac{16\,(y^{2}+4)}{x^{2}+4}\), all factors cancel except -16x.

∴  The value of the given expression is  \(-16x\).

Hence, the correct answer is Option 3. 

Calculation:

Given,

\(x = \sec\theta - \cos\theta\)

\(y = \sec^4\theta - \cos^4\theta\)

Compute derivatives w.r.t. \(\theta\):

\(\dfrac{dx}{d\theta} = \sec\theta\,\tan\theta + \sin\theta\)

\(\dfrac{dy}{d\theta} = 4\sec^4\theta\,\tan\theta \;+\; 4\cos^3\theta\,\sin\theta\)

From the ratio \(\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)

\(\dfrac{dy}{dx} = \frac{4\bigl(\sec^4\theta\,\tan\theta + \cos^3\theta\,\sin\theta\bigr)} {\sec\theta\,\tan\theta + \sin\theta}\)

Using identities \(x^2 = \sec^2\theta + \cos^2\theta - 2\), \(\sec^3\theta + \cos^5\theta)^2 = y^2 + 4\), and \((1 + \cos^2\theta)^2 = x^2 + 4,\)

\(\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = 16 \,\frac{\,y^{2} + 4\,}{\,x^{2} + 4\,} \)

∴  \(\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = 16\,\frac{y^{2} + 4}{x^{2} + 4} \)

Hence, the correct answer is Option 3.

Concept:

a² - b² = (a - b) (a + b)

sec x = 1/cos x and cosec x = 1/sin x

a³ + b³ = (a + b) (a² + b² - ab)

Calculation:

 \(\frac{{\left( {1 - {\rm{sinAcosA}}} \right)\left( {{\rm{si}}{{\rm{n}}^2}{\rm{A}} - {\rm{co}}{{\rm{s}}^2}{\rm{A}}} \right)}}{{{\rm{cosA}}\left( {{\rm{secA}} - {\rm{cosecA}}} \right)\left( {{\rm{si}}{{\rm{n}}^3}{\rm{A}} + {\rm{co}}{{\rm{s}}^3}{\rm{A}}} \right)}}\) 

⇒ \( \frac{{\left( {{\rm{1}} - {\rm{sinAcosA}}} \right)\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {{\rm{sinA}} - {\rm{cosA}}} \right)}}{{{\rm{cosA}}\left[ {\frac{1}{{{\rm{cosA}}}} - \frac{1}{{{\rm{sinA}}}}} \right]\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {{\rm{si}}{{\rm{n}}^2}{\rm{A}} + {\rm{co}}{{\rm{s}}^2}{\rm{A}} - {\rm{sinAcosA}}} \right)}}\)

⇒ \( \frac{{\left( {1 - {\rm{sinAcosA}}} \right)\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {{\rm{sinA}} - {\rm{cosA}}} \right)}}{{{\rm{cosA}}\left[ {\frac{{{\rm{sinA}} - {\rm{cosA}}}}{{{\rm{sinA}}.{\rm{cosA}}}}} \right]\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {1 - {\rm{sinAcosA}}} \right)}}\)

⇒ \(\frac{sinA - cosA}{cosA[\frac{sinA - cosA}{sinA.cosA}]}\)

⇒ \(\frac{(sinA - cosA)\times sinA.cosA}{cosA[sinA - cosA]}\)

⇒ \(\frac{ sinA.cosA}{cosA}\)

⇒ sin A

∴ The correct answer is option (1).

Concept:

cos 2x = cos² x – sin² x = 2 cos² x - 1 = 1 – 2 sin² x

Calculation:

cos 4x

= cos 2(2x)

= 2 cos² 2x – 1                                         (∵cos 2x = 2 cos² x – 1)

= 2 (1 – 2 sin² x)² – 1                              (∵ cos 2x = 1 – 2 sin² x)

= 2 [1 – 4 sin² x + 4 sin⁴ x] – 1               [∵ (a – b)² = a² – 2ab + b²]

= 2 - 8 sin² x + 8 sin⁴ x – 1

= 1 - 8 sin² x + 8 sin⁴ x

Concept:

If sin θ = sin α then θ = nπ + (- 1)ⁿ α, α ∈ [-π/2, π/2], n ∈ Z.

Calculation:

Given: 4 sin 3x = 2

⇒ sin 3x = ½

As we know that, sin (π/6) = ½

⇒ sin 3x = sin (π/6)

As we know that, if sin θ = sin α then

θ = nπ + (- 1)ⁿ α, α ∈ [-π/2, π/2], n ∈ Z.

⇒ 3x = nπ + (- 1)ⁿ × (π/6), where n ∈ Z.

Concept:

General solution of some standard trigonometric equations:

Calculation:

Given: cos x = 1

⇒ cos x = cos 0

As we know, If cos θ = cos α then θ = 2nπ ± α

Therefore, x = 2nπ ± 0 = 2nπ 

Hence, the general solution of cos x = 1 is x = 2nπ, n ∈ Z.

Concept:

sin2x + cos2x = 1

Calculations:

Given, 4sin2x -  2 cos2x = 2

⇒2sin2x -  cos2x = 1

⇒ 2sin2x -  cos2x = sin2x + cos2x

⇒ sin2x = 2 cos2x

⇒tan2 x = 2

⇒ tan x = √2

Concept:

\(\rm cot (A + B) = \dfrac{cot A cot B - 1}{cot A + cot B}\)

Calculation:

Given,

(cot 2x ∙ cot 4x) - (cot 4x ∙ cot 6x) - (cot 6x ∙ cot 2x)

⇒ (cot 2x ∙ cot 4x) - cot 6x [cot 4x + cot 2x]

⇒ (cot 2x ∙ cot 4x) - cot (2x + 4x) [cot 4x + cot 2x]

⇒ (cot 2x ∙ cot 4x) -  \((\rm \dfrac{cot 2x .cot 4x - 1}{cot 2x + cot 4x})\)× [cot 4x + cot 2x]

⇒ (cot 2x ∙ cot 4x) - (cot 2x ∙ cot 4 x - 1)

⇒ (cot 2x ∙ cot 4x) - (cot 2x ∙ cot 4x) + 1

⇒ 1

Concept:

\(\rm \sin \theta = \frac{Perpendicular}{Hypotenuse}\)

\(\rm \cos \theta = \frac{Base}{Hypotenuse} \)

sin2 θ + cos2 θ = 1 

Calculation:

7 sinθ + 24 cosθ = 25

Dividing by 25 on both the sides, we get

\(\rm \frac{7}{25}\)sinθ + \(\rm \frac{24}{25}\)cosθ = 1      ....(i)

We know that,

sin² θ + cos² θ = 1  

sin θ.sin θ + cos θ.cos θ = 1      ....(ii)

On comparing equ (i) and (ii)

sin θ = \(\rm \frac{7}{25}\) 

cos θ = \(\rm \frac{24}{25}\)

Now, (sinθ + cosθ) 

= \(\rm \frac{7}{25}\)+ \(\rm \frac{24}{25}\)

= \(\rm \frac{31}{25}\)

Concept:

cos (A + B) = cosA. cosB - sinA. sin B

Calculation:

Given: cos 2B = 3 sin² A   .... (1)

and  3sin 2A = 2 sin 2B     .... (2)

Divide (1) by (2),

\(\frac{cos(2B)}{2sin(2B)}=\frac{3sin^2(A)}{3sin(2A)}\) .... (3)

We know that,

sin 2A = 2 sin A∙ cos A

Hence, equation (3) becomes,

\(\frac{cos(2B)}{2sin(2B)}=\frac{sin^2(A)}{2sin(A)cos(A)}\)

⇒ \(\frac{cos(2B)}{sin(2B)}=\frac{sin(A)}{cos(A)}\)

⇒ cos (2B) cos (A) = sin (2B) sin (A)

⇒ cos (2B) cos (A) - sin (2B) sin (A) = 0

From above concept,

cos (A + 2B) = 0

⇒ (A + 2B) = cos^-1 (0)

⇒ (A + 2B) =  π/2

Concept:

If tan θ = tan α then θ = nπ + α, α ∈ (-π/2, π/2), n ∈ Z.

Calculation:

Given: 8 tan(2x) – 5 = 3

⇒ 8 tan(2x) = 8

⇒ tan 2x = 1

As we know that, tan (π/4) = 1

⇒ tan 2x = tan (π/4)

As we know that, if tan θ = tan α then θ = nπ + α, α ∈ (-π/2, π/2), n ∈ Z.

⇒ 2x = nπ + (π/4), where n ∈ Z.

Concept:

2sinx cos x = sin 2x

sin (90° + x) = cos x

Calculation:

Given, 2 sin 75° cos 75°

= sin [2 (75°)]

= sin (150°)

= sin (90° + 60°)

= cos 60°

= \(\rm \dfrac 1 2\)

Hence, the value of 2 sin 75° cos 75° is \(\rm \dfrac 1 2\)