Question

If cosƟ+cos2Ɵ =1,the value of sin2Ɵ+sin4Ɵ is
(a) -1

(b) 0

(c) 1

(d) 2​

Answer 1

Answer:

b

Step-by-step explanation:

Given

$\cos(x) + \cos(2x) = 1 \\ \\ \implies \: \cos(x) + 2 \cos {}^{2}(x ) - 1 = 1 \\ \\ let \: \cos(x) = y \\ \\ \implies 2 {y}^{2} + y - 1 = 0 \\ \implies 2 {y}^{2} + 2y - y - 1 \\ \implies (2y - 1)(y + 1) = 0 \\ \\ \implies y = \frac{1}{2} \: \: or \: \: - 1 \\ \\ \cos(x) = \frac{1}{2} \: \: or \: \: - 1 \\ \\ x = \frac{\pi}{3} \: \: or \: \: \pi$

So now put value in required expression

$when \: x = \frac{\pi}{3} \\ \\ \sin(2x) + \sin(4x) \\ \\ = \sin({2\pi \over3}) + \sin( \frac{4\pi}{3} ) \\ \\ = \frac{ \sqrt{3} }{2} + \frac{ \sqrt{3} }{2} = \sqrt{3}$

Or When x = π then

$\sin(2x) + \sin(4x) \\ = \sin(2\pi) + \sin(4\pi) \\ = 0 + 0 = 0$

Answer 2

SOLUTION

GIVEN

$\sf{ \cos \theta + { \cos}^{2} \theta = 1}$

TO CHOOSE THE CORRECT OPTION

$\sf{ { \sin}^{2} \theta +{ \sin}^{4} \theta } =$

(a) - 1

(b) 0

(c) 1

(d) 2

EVALUATION

Here it is given that

$\sf{ \cos \theta + { \cos}^{2} \theta = 1}$

Now

$\sf{ \cos \theta + { \cos}^{2} \theta = 1}$

$\sf{ \implies \cos \theta = 1 - { \cos}^{2} \theta }$

$\sf{ \implies \cos \theta = { \sin}^{2} \theta }$

Now

$\sf{ { \sin}^{2} \theta +{ \sin}^{4} \theta }$

$\sf{ = { \sin}^{2} \theta + {({ \sin}^{2} \theta)}^{2} }$

$\sf{ = { \sin}^{2} \theta + {( \cos \theta)}^{2} }$

$\sf{ = { \sin}^{2} \theta + {\cos}^{2} \theta}$

$\sf{ = 1 }$

FINAL ANSWER

Hence the correct option is (c) 1

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