Question

(Multiple choice trig question) How do I evaluate cos(x)sin(x-90) + sin^2(x) ?

Answer 1

A) 0,
B) 1,
C) 2x,
D) 2sin(x)

Answer 2

use the identity \[\large\rm \sin(a+b)= \sin A *cosB - \cos A \sin B\]
for sin(x-90)

Answer 3

@Nnesha do I set it up as
cos(x)sin(a + b) ?

Answer 4

\[\huge\rm cos(x)\color{ReD}{sin(x-90)} + \sin^2(x)\]
just for red part use that formula i gave you

sin(x-90) a=x
and b = 90

Answer 5

i see i meant to say sin(a-b) not a+B

Answer 6

sorry about that let me retype it

Answer 7

use the identity \[\large\rm \sin(a-b)= \sin A *cosB - \cos A \sin B\]
for sin(x-90)
where
a= x
and b =90

Answer 8

so sin(x-90)=??

Answer 9

@Nnesha
sin(x-90) = -cos(x)

Answer 10

hmm right \[\huge\rm cos(x)*\color{ReD}{-cos(x)} + \sin^2(x)\]

multiply cos(x) b -cos(x)

Answer 11

you're familiar with cos^2x + sin^2 x=1 identity solve this equation for sin^2(x)

Answer 12

@Nnesha I have
cos(x)(-cos(x)+sin^2
-cos^2(x)+sin^2(x)
-cos(2x)

Answer 13

hmm 2nd line is correct
use the identity to substitute sin^2(x)
\[\rm \sin^2\theta+\cos^2 \theta =1\]
solve this identity for sin^2(x)

Answer 14

\[\rm \sin^2\theta+\cos^2 \theta =1\] \[\sin^2 \theta = ???\]

Answer 15

@Nnesha using the identity I have

cos^2(x)+sin^2(x)+(-cos(x))cos(x)
cos^2(x)+cos(x)(-cos(x)+sin^2(x)
-cos^2(x)+cos^2(x)+sin^2(x)
sin^2(x) + 0
sin^2(x)

Answer 16

hmm the first identity we used should be in form
\[\sin(x-90)= \sin x * \cos 90 - \cos x *\sin 90 \]

Answer 17

how did you get -cos(x) and cos(x) at first line ?

Answer 18

were you talking about this identity ???
\[\rm \sin^2\theta+\cos^2 \theta =1\]

Answer 19

@Nnesha yes

Answer 20

okay
sin^2 x + cos^2 x =1
subtract cos^2 x both sides to solve for sin^2x

Answer 21

what do you get ? let m know ))

Answer 22

@Nnesha
sin^2 x + cos^2 x =1
sin^2x = -sin^2x
-1 ?

Answer 23

subtract both sides you will get \[\large\rm \sin^2x= 1-\cos^2x\]