Question

Question #8bd05

Answer 1

See the Explanation.

Explanation:

Recall that, `costheta=sin(90^@-theta)..........(1)`

`theta=60^@+x rArr90^@-theta=90^@-(60^@+x)=30^@-x.`

`:.," by "(1), cos(60^@+x)=sin(30^@-x).`

Hence, the Verification.

Answer 2

Proved.

Explanation:

Prove: `cos(60^@ +x)=sin(30^@-x)`

Use the identity `cos(A+B) = cos(A)cos(B)-sin(A)sin(B)` where `A = 60^@` and `B = x` on the left side:

`cos(60^@)cos(x)+sin(60^@)sin(x)=sin(30^@-x)`

Use the fact `cos(60^@) = sin(30^@)` on the left side:

`sin(30^@)cos(x)+sin(60^@)sin(x)=sin(30^@-x)`

Use the fact `sin(60^@) = cos(30^@)` on the left side:

`sin(30^@)cos(x)+cos(30^@)sin(x)=sin(30^@-x)`

Substitute -x for x on the left side:

`sin(30^@)cos(-x)+cos(30^@)sin(-x)=sin(30^@-x)`

Use the fact that the cosine function is even (`cos(-x) = cos(x)`) on the left side:

`sin(30^@)cos(x)+cos(30^@)sin(-x)=sin(30^@-x)`

Use the fact that the sine function is odd (`sin(-x) = -sin(x)`) on the left side:

`sin(30^@)cos(x)-cos(30^@)sin(x)=sin(30^@-x)`

Use the identity `sin(A-B) = sin(A)cos(B)-cos(A)sin(B)` where `A = 30^@` and `B = x` on the left side:

`sin(30^@-x)=sin(30^@-x)`

Q.E.D.

Answer 3

see explanation.

Explanation:

Simplify both sides of the identity and compare.

Using the following `color(blue)"Addition formulae"`

`• sin(A-B)=sinAcosB-cosAsinB`

`• cos(A+B)=cosAcosB-sinAsinB`

`"left side " = cos(60+x)^@`

`color(white)(left side)=cos60^@cosx^@-sin60^@sinx^@`

`color(white)(left side)=1/2cosx^@-sqrt3/2sinx^@larr" left side"`

`color(orange)"Reminder" [cos60^@=sin30^@=1/2]`

`"and " [cos30^@=sin60^@=sqrt3/2]`

`"right side " =sin(30-x)^@`

`color(white)(right side)=sin30^@cosx^@-cos30^@sinx^@`

`color(white)(right side)=1/2cosx^@-sqrt3/2sinx^@larr" right side"`

`"left side " =" right side "rArr" verified"`