`sin(t) = 4/5 a) sin(pi - t) b) sin(t + pi)` Use the value of the trigonometric function to evaluate the indicated functions.

Textbook Question

Chapter 4, 4.2 - Problem 41 - Precalculus (3rd Edition, Ron Larson).
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kspcr111 | In Training Educator

Posted on

Given

`sin(t) = 4/5`

a) `sin(pi - t)`

this is of the form sin(a-b) where `a= pi` and `b= t`

`sin(a-b) = sin(a)cos(b) - cos(a)sin(b)`

so,

`sin(pi - t)=sin(pi)cos(t) - cos(pi)sin(t)` =` 0 - (-1)sin(t) = sin(t)`

or simply we know by the identity `sin(pi-t) = sin(t)` we get

`sin(pi-t) = sin(t) = 4/5 `

b) `sin(t + pi)`

similar to above we get

`sin(t + pi)= sin(pi+t) = -sin(t)`

we can also get this by the identity 

so now ,

`sin(t + pi) = - sin(t) = -(4/5)`

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