Question

Given that `sin theta cos theta =7/50` and `0^0 < theta < 90^0`. Find the value of `cos theta`.?

Help pls...

Answer 1

`7/(5*sqrt(2)),1/(5*sqrt(2))`

Explanation:

Solving the equation
`sin(theta)*cos(theta)=7/50`
in the given interval we get

`x_1=-2arctan(7-5sqrt(2))`

`x_2=2arctan(1/7(5sqrt(2)-1))`
so we gat

`cos(x_1)=7/(5sqrt(2))`

`cos(x_2)=1/(5sqrt(2))`

Answer 2

`8^@13, 81^@87`

Explanation:

`sin t.cos t = 7/50`
Use trig identity: sin 2t = 2sin t.cos t
In this case:
`sin t.cos t = (sin 2t)/2 = 7/50`
`sin 2t = 15/50 = 7/25`
Calculator and unit circle give 2 solutions for 2t:
`2t = 16^@26`, and `2t = 180 - 16.26 = 163^@74`
a. 2t = 16.26 + k360
`t = 8^@13 + k180^@`
b. 2t = 163.74 + k360
`t = 81^@87 + k180^@`
Answers for (0, 90)
`8^@13, 81^@87`
cos (8.13) = 0.99
cos (81.87) = 0.14