Question

For what value of x is sin x = cos 19°, where 0°< x < 90°?

Answer 1

Since u have asked sin x=1/2
According to identity,
Sin2 x +cos2 x =1
Cos 2 x =1 − sin 2x
Cos x = √1−sin 2x
Cos x =√1−(1/2)2
Cos x =√1−1/4
Cos x =√3/4
Cos x =√3/2
By this way u can find the cos of given sin and vice versa because because it is not possible that we always know the value of given trigonometric function orally eg. Sin x = 1/10 etc…….
Hope u understand what i want to say

Answer 2

TA DA

Answer 3

!!!!!

Answer 4

bye

Answer 5

no i dont

Answer 6

hint: \(\bf \textit{Cofunction Identities}
\\ \quad \\
sin\left(\frac{\pi}{2}-{\color{red}{ \theta}}\right)=cos({\color{red}{ \theta}})\qquad

cos\left(\frac{\pi}{2}-{\color{red}{ \theta}}\right)=sin({\color{red}{ \theta}})\)

any ideas?

Answer 7

or lemme put it differently \(\bf sin\left(90^o-{\color{red}{ \theta}}\right)=cos({\color{red}{ \theta}})\qquad

cos\left(90^o-{\color{red}{ \theta}}\right)=sin({\color{red}{ \theta}})
\)