Question

Find the value of, sin(45°+x)-cos(45°-x)

Answer 1

Answer:

Value of

Value of sin(45°+x)-cos(45°-x) = 0

Step-by-step explanation:

sin(45°+x)-cos(45°-x)

=cos[90°-(45°+x)]-cos(45°-x)

/* Since, cos(90-A) = sinA */

= cos(90°-45°-x)-cos(45°-x)

= cos(45°-x)-cos(45°-x)

= 0

Therefore,

Value of

sin(45°+x)-cos(45°-x) = 0

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Answer 2

Answer:  The required value of the given expression is 0.

Step-by-step explanation:

We have the following trigonometric formulae :

$(i)~\sin(A+B)=\sin A\cos B+\cos A\sin B,\\\\(ii)~\cos(A-B)=\cos A\cos B+\sin A\sin B.$

Using the above formulae in the given expression, we get

$\sin(45^\circ+x)-\cos(45^\circ-x)\\\\=(\sin45^\circ \cos x+\cos 45^\circ \sin x)-(\cos45^\circ \cos x+\sin 45^\circ \sin x)\\\\=\dfrac{1}{\sqrt2}\cos x+\dfrac{1}{\sqrt2}\sin x-\dfrac{1}{\sqrt2}\cos x-\dfrac{1}{\sqrt2}\sin x~~~~[\textup{since }\sin45^\circ=\cos45^\circ=\dfrac{1}{\sqrt2}]\\\\\\=0.$

Thus, the required value of the given expression is 0.