Question

sin (45° + θ) – cos (45° – θ) is equal to

2 cos θ

0

2 sin θ

1

​

Answer 1

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

Answer 2

Answer:

★

sin (45 + theta) - Cos (45 - theta)

=sin(45 + O) – cos(45 – O)

=sin(45 + O) – sin{90- (45 – O)} (since, cosO =sin (90 – O))

= sin(45 + O) – sin{45 + O}

=0