UPSC NDA Math Model Paper 8

Let us consider B = 18°
So, 5B = 90°
⇒ 2B + 3B = 90˚
⇒ 2B = 90˚ - 3B
By taking sine on both sides, we get
Sin 2B = sin (90˚ - 3B) = cos 3B
⇒2‌sin‌B‌cos‌B=4cos3B−3‌cos‌B
⇒2‌sin‌B‌cos‌B−4cos3B+3‌cos‌B=0
⇒cos‌B(2‌sin‌B−4cos2B+3)=0
Dividing both sides by cos‌B=cos‌18°≠0, we get
⇒2‌sin‌B−4(1−sin2B)+3=0
⇒4sin2B+2‌sin‌B−1=0
This is a quadratic equation in Sine,
So, sin‌B=

−2±√4+16

2×4

=

−1±√5

4

Now since 18° is first quadrant so sin 18° is positive
Therefore, sin‌B=

−1+√5

4

Now, cos‌36°=cos(2×18°)
⇒cos‌36°=1−2sin218°
⇒cos‌36°=1−2(

√5−1

4

)2 =(

16−2×(5+1−2√5

16

)=

4+4√5

16

Cos‌36°=

√5+1

4