1.

Sin (45° + θ) – cos (45° – θ) is equal to2 cos θ02 sin θ1​

Answer»

QUESTION :

Find the VALUE of:

sin(45° + θ) – cos (45°-θ)

Formula's USED :

\sf1)\sin(a+b)=\sin\:a\cos\:b+\sin\:b\cos\:a

\sf2)\sin(a-b)=\sin\:a\cos\:b-\sin\:b\cos\:a

\sf3)\cos(a+b)=\cos\:a\cos\:b-\sin\:b\sin\:a

\sf4)\cos(a-b)=\cos\:a\cos\:b+\sin\:b\sin\:a

Solution :

We have ,

sin(45° + θ) – cos (45°-θ)

USE formula of sin(a+b) and cos(a-b) ,then

sin(45° + θ) – cos (45°-θ)

=sin45°cosθ+sinθcos45°-(cosθcos45°-sinθsin45°)

=sin45°cosθ+sinθcos45°-cosθcos45°+sinθsin45°

We know that cos45°=sin45°

=sin45°cosθ+sinθcos45°-cosθsin45°+sinθcos45°

=0

Therefore,sin(45° + θ) – cos (45°-θ)=0

Another WAY to solve :

We know that sin(90-x)=cosx and cos(90-x)=sinx

Then,

sin(45° + θ) – cos (45°-θ)

=sin(45° + θ) – sin[90-(45°-θ)]

=sin(45° + θ) – sin(45°+θ)

=0


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