CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$\cos x + \cos^{2}x = 1$$, then the value of $$\sin^{4}x + \sin^{6}x$$ is


A
1+5
loader
B
152
loader
C
152
loader
D
1+52
loader
E
15
loader

Solution

The correct option is E $$\dfrac {-1 + \sqrt {5}}{2}$$
Given, $$\cos x + \cos^{2}x = 1$$     .... (i)
$$\Rightarrow \cos^{2}x + \cos x - 1 = 0$$
Therefore, $$\cos x = \dfrac {-1\pm \sqrt {1 + 4}}{2} (\because$$ quadratic in $$\cos x)$$
$$= \dfrac {-1\pm \sqrt {5}}{2} = \dfrac {-1 + \sqrt {5}}{2} \left (\because \cos x \neq \dfrac {-1 - \sqrt {5}}{2}\right )$$
Also from Eq. (i),
$$\cos x = 1 - \cos^{2} x$$
$$\Rightarrow \cos x = \sin^{2}x$$     .... (ii)
Now, $$\sin^{4}x + \sin^{6}x$$
$$= \cos^{2} x + \cos^{3}x$$
$$= \cos^{2} x (1 + \cos x)$$
$$= \left (\dfrac {-1 + \sqrt {5}}{2}\right )^{2} \left (1 + \dfrac {-1 + \sqrt {5}}{2}\right )$$
$$= \left (\dfrac {-1 + \sqrt {5}}{2}\right )^{2} \left (\dfrac {1 + \sqrt {5}}{2}\right )$$
$$= \left (\dfrac {-1 + \sqrt {5}}{2}\right ) \left (\dfrac {5 - 1}{4}\right ) = \left (\dfrac {-1 + \sqrt {5}}{2}\right ) \times 1$$
$$= \dfrac {-1 + \sqrt {5}}{2}$$

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image