Question

2sin(45+theta)sin(45-theta)÷cos2theta

Answer 1

The given expression is

$\frac{2\sin (45^o+\theta)\sin (45^o-\theta)}{\cos (2\theta)}$

Now using the product formula,

$2\sin (A+B)\sin (A-B)=\cos (2B)-\cos (2A)$

The above expression is

$\frac{2\sin (45^o+\theta)\sin (45^o-\theta)}{\cos (2\theta)} = \frac{\cos (2\theta)-\cos (90^o)}{\cos (2\theta)} =\frac{\cos (2\theta)-0}{\cos (2\theta)} =1$

Answer 2

Answer:

The given expression is

\frac{2\sin (45^o+\theta)\sin (45^o-\theta)}{\cos (2\theta)}

cos(2θ)

2sin(45

o

+θ)sin(45

o

−θ)

Now using the product formula,

2\sin (A+B)\sin (A-B)=\cos (2B)-\cos (2A)2sin(A+B)sin(A−B)=cos(2B)−cos(2A)

The above expression is

\frac{2\sin (45^o+\theta)\sin (45^o-\theta)}{\cos (2\theta)} = \frac{\cos (2\theta)-\cos (90^o)}{\cos (2\theta)} =\frac{\cos (2\theta)-0}{\cos (2\theta)} =1

cos(2θ)

2sin(45

o

+θ)sin(45

o

−θ)

=

cos(2θ)

cos(2θ)−cos(90

o

)

=

cos(2θ)

cos(2θ)−0