If sin(x) = 1/5 and x is in quadrant I, find the exact values of the expressions without solving for x.

Question

If sin(x) = 1/8 and x is in quadrant I, find the exact values of the expressions without solving for x.

A) sin(2x)

B) cos(2x)

C) tan(2x)

Answer 1

Let's continue finding tan(2x).

We earlier stated in the video about the formula tan(2x) = ( 2 * tan(x) ) / ( 1 - tan^2(x) ).

Tangent is defined as tan(x) = opposite / adjacent, so for this problem, tan(x) = sqrt(6) / 12.

Plugging in appropriate numbers, tan(2x) = ( sqrt(6) / 6) / ( 1 - 6/144 ) = 12 * sqrt(6) / 19 ....... (ANS).

The exact same process can be used to find when sin(x) = 1/8. Create a triangle, and identify 1 as the vertical length, 8 as the hypotenuse, and by Pythagorean theorem, adjacent = sqrt( 8^2 - 1^2) = sqrt(63).

So, sin(x) = 1/8. cos(x) = sqrt(63) / 8, and tan(x) = sqrt(63) / 63 = sqrt(7) / 21 (radicalized it).

Therefore, sin(2x) = 2 * 1/8 * sqrt(63) / 8 = 3 * sqrt(7) / 32

cos(2x) = 2 * ( sqrt(63) / 8 )^2 - 1 = 31/32

tan(2x) = 2 * ( sqrt(7) / 21 ) / ( 1 - ( sqrt(7) / 21 )^2 ) = 3 * sqrt(7) / 31 >>>>> (END)

Hope this helps! Let me know in the comments of any feedback!