UPSC NDA Math Model Paper 6

Given: sin‌2‌θ=cos‌3‌θ,0<θ<π∕2.
As we know that, sin‌2‌θ=2‌sin‌θ‌cos‌θ and cos‌3‌θ=4cos3θ−3‌cos‌θ.
⇒2‌sin‌θ‌cos‌θ=4cos3θ−3‌cos‌θ
⇒2‌sin‌θ=4cos2θ−3
⇒2‌sin‌θ=4(1−sin2θ)−3=4−4sin2θ−3
⇒4sin2θ+2‌sin‌θ−1=0
Comparing the above equation with quadratic equation ax2+bx+c=0,a=4,b=2 and c=−1
Now substituting the values in the quadratic formula x=(−b±√b2−4ac)∕2a we get,
sin‌θ‌=‌

−2±√−22−4(4)(−1)

2(4)

=

−2±√4+16

8

=

−2±√20

8

=

−1±√5

4

Thus, sin‌θ=

−1±√5

4

.
Since, 0 < θ < π/2 ⟹ θ lies between 0° to 90°⟹ all ratios are positive.
⇒sin‌θ=

−1+√5

4

As we know that, cos2θ=cos2θ−sin2θ=1–2sin2θ
⇒cos‌2‌θ=‌1−2(

−1+√5

4

)2=

1+√5

4