JEE maths #1

Geometry Level 4

If f ( x ) = 2 sin ⁡ 2 θ + 4 cos ⁡ ( x + θ ) sin ⁡ x sin ⁡ θ + cos ⁡ ( 2 x + 2 θ ) , f(x) = 2\sin^{2}\theta + 4\cos(x+\theta)\sin x \sin \theta + \cos(2x+2\theta), find the value of f 2 ( x ) + f 2 ( π 4 − x ) . f^{2}(x) + f^{2}\left(\frac \pi 4-x\right).

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The answer is 1.

Rahil Sehgal by Rahil Sehgal , 20, India

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2 solutions

Prakhar Bindal
Mar 29, 2017

As the problem is for jee . i will give a solution for examination

As answer is independent of theta . Put theta = 0

hence f(x) = cos2x

Hence given expression is cos^2(2x)+sin^2(2x) = 1

11 Helpful 0 Interesting 0 Brilliant 0 Confused

Very nice and new way to solve such type of problem... Upvoted✓ :)

Rahil Sehgal - 4 years, 2 months ago

You could have done it even faster if you put theta and x equal to zero

Sabhrant Sachan - 4 years, 2 months ago

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U cannot directly put the value of x =0.. Only θ = 0 \theta= 0 is possible as θ \theta is an independent term

Rahil Sehgal - 4 years, 2 months ago

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θ = 0 , x = 0 \theta =0 , x =0 .

f ( 0 ) = 1 f(0) = 1

θ = 0 , x = π 4 \theta =0 , x = \dfrac{\pi}{4}

f ( π 4 ) = 0 f\left(\dfrac{\pi}{4}\right) = 0

f 2 ( 0 ) + f 2 ( π 4 ) = 1 f^2(0)+ f^2\left(\dfrac{\pi}{4}\right) = 1

Sabhrant Sachan - 4 years, 2 months ago

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@Sabhrant Sachan – Well i didn't need a pen even to solve . cos^2(x)+sin^2(x) = 1 is a 10th class identity :P

Prakhar Bindal - 4 years, 2 months ago

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@Prakhar Bindal – I agree :) XD

Sabhrant Sachan - 4 years, 2 months ago

@Sabhrant Sachan – Thanks for your explanation...

Rahil Sehgal - 4 years, 2 months ago

HeeHee we JEE aspirants often do it. And i also did it.

Nivedit Jain - 3 years, 12 months ago

Old trick boi B-)

Sahil Silare - 3 years, 2 months ago

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Yeah! Now it seems to be old unlike before.

Rahil Sehgal - 3 years, 2 months ago
Chew-Seong Cheong
Mar 31, 2017

f ( x ) = 2 sin ⁡ 2 θ + 4 cos ⁡ ( x + θ ) sin ⁡ x sin ⁡ θ + cos ⁡ ( 2 x + 2 θ ) Note: sin ⁡ A sin ⁡ B = 1 2 ( cos ⁡ ( A − B ) − cos ⁡ ( A + B ) ) = 2 sin ⁡ 2 θ + 2 cos ⁡ ( x + θ ) ( cos ⁡ ( x − θ ) − cos ⁡ ( x + θ ) ) + 2 cos ⁡ 2 ( x + θ ) − 1 Note: 2 cos ⁡ 2 A = cos ⁡ 2 A − 1 = 2 sin ⁡ 2 θ + 2 cos ⁡ ( x + θ ) cos ⁡ ( x − θ ) − 1 Note: 2 cos ⁡ A cos ⁡ B = cos ⁡ ( A − B ) + cos ⁡ ( A + B ) = 2 sin ⁡ 2 θ + cos ⁡ ( 2 x ) + cos ⁡ ( 2 θ ) − 1 = 2 sin ⁡ 2 θ + cos ⁡ ( 2 x ) + 2 cos ⁡ 2 θ − 1 − 1 = cos ⁡ ( 2 x ) \begin{aligned} f(x) & = 2\sin^2 \theta + 4\cos(x+\theta){\color{#3D99F6}\sin x \sin \theta} + {\color{#D61F06}\cos(2x+2\theta)} & \small \color{#3D99F6} \text{Note: }\sin A \sin B = \frac 12 (\cos(A-B) - \cos (A+B)) \\ & = 2\sin^2 \theta + 2\cos(x+\theta)({\color{#3D99F6}\cos(x-\theta) - \cos(x+ \theta )}) + {\color{#D61F06}2\cos^2(x+\theta)-1} & \small \color{#D61F06} \text{Note: } 2\cos 2A = \cos^2 A - 1 \\ & = 2\sin^2 \theta + {\color{#3D99F6}2 \cos(x+\theta) \cos(x- \theta)} -1 & \small \color{#3D99F6} \text{Note: } 2\cos A \cos B = \cos(A-B) + \cos (A+B) \\ & = 2\sin^2 \theta + {\color{#3D99F6} \cos(2x) + \cos(2\theta)} -1 \\ & = 2\sin^2 \theta + \cos(2x) + 2\cos^2 \theta - 1 -1 \\ & = \cos (2x) \end{aligned}

Therefore, we have:

f 2 ( x ) + f 2 ( π 4 − x ) = cos ⁡ 2 ( 2 x ) + cos ⁡ 2 ( π 2 − 2 x ) = cos ⁡ 2 ( 2 x ) + sin ⁡ 2 ( 2 x ) = 1 \begin{aligned} f^2(x) + f^2\left(\frac \pi 4- x\right) & = \cos^2 (2x) + \cos^2 \left(\frac \pi 2- 2x\right) \\ & = \cos^2 (2x) + \sin^2 (2x) \\ & = \boxed{1} \end{aligned}

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