# Calculate the value of the sum sin(arcsin(1/4))+cos(2arccos(1/4)).

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### 2 Answers

We have to find the value of sin(arc sin(1/4)) + cos(2*arc cos(1/4))

sin(arc sin(1/4)) + cos(2*arc cos(1/4))

=> sin ( arc sin (1/4)) + 2*(cos (arc cos(1/4))^2 - 1

=> 1/4 + 2*(1/4)^2 - 1

=> 1/4 + 2*(1/16) - 1

=> 1/4 + 1/8 - 1

=> -5/8

**The required value of the sum is -5/8.**

We'll start from the fact that sin(arcsin x) = x.

Comparing, we'll get:

sin (arcsin(1/4)) = 1/4

We'll note arccos(1/4) = a

cos (2arccos(1/4)) = cos 2a

We'll apply the double angle identity:

cos 2a = 2(cos a)^2 - 1

If a = arccos(1/4) => (cos a)^2 = (cos arccos(1/4))^2 = 1/4^2

The sum will become:

S = 1/4 + 2/4^2 - 1

S = 1/4 + 2/16 - 1

S = (4+2-16)/16

S = -10/16

S = -5/8

**The value of the trigonometric sum sin(arcsin(1/4))+cos(2arccos(1/4)) = -5/8.**