Question

Find the exact value by using a half-angle identity. sine of five pi divided by twelve

Answer 1

Answer : The exact value of $\sin (\frac{5\pi}{12})=\frac{\sqrt{(2+\sqrt{3})}}{2}$

Step-by-step explanation :

As we are given that: $\sin (\frac{5\pi}{12})$

Using a half-angle identity:

$\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos\theta}{2}}$

$\sin\left(\frac{\theta}{2}\right)=\sin \frac{(\frac{5\pi}{6})}{2}$

Thus, $\theta=\frac{5\pi}{6}$

Now using half-angle identity, we get:

$\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos\theta}{2}}$

$\sin\left(\frac{(\frac{5\pi}{6})}{2}\right)=\sqrt{\frac{1-\cos (\frac{5\pi}{6})}{2}}$

$\sin\left(\frac{(\frac{5\pi}{6})}{2}\right)=\sqrt{\frac{1-\cos (\pi -\frac{\pi}{6})}{2}}$

$\sin\left(\frac{(\frac{5\pi}{6})}{2}\right)=\sqrt{\frac{1+\cos (\frac{\pi}{6})}{2}}$

As we know that,

$\cos \frac{\pi}{6}=\cos 30^o=\frac{\sqrt{3}}{2}$

Now put the value of $\cos \frac{\pi}{6}$, we get:

$\sin (\frac{5\pi}{12})=\sqrt{\frac{1+(\frac{\sqrt{3}}{2})}{2}}$

$\sin (\frac{5\pi}{12})=\sqrt{\frac{(\frac{2+\sqrt{3}}{2})}{2}}$

$\sin (\frac{5\pi}{12})=\sqrt{(\frac{2+\sqrt{3}}{4})}$

$\sin (\frac{5\pi}{12})=\frac{\sqrt{(2+\sqrt{3})}}{2}$

Thus, the exact value of $\sin (\frac{5\pi}{12})=\frac{\sqrt{(2+\sqrt{3})}}{2}$

Answer 2

Answer:

cos

(

5

π

12

)

=

√

2

−

√

3

2

Explanation:

By the half angle formula:

XXXX

cos

(

θ

2

)

=

±

√

1

+

cos

(

θ

)

2

If  

θ

2

=

5

π

12

XXXX

then  

θ

=

5

π

6

Note that  

5

π

6

is a standard angle in quadrant 2 with a reference angle of  

π

6

so  

cos

(

5

π

6

)

=

−

cos

(

π

6

)

=

−

√

3

2

Therefore

XXXX

cos

(

5

π

12

)

=

±







⎷

1

−

√

3

2

2

XXXXXXXXXXX

=

±







⎷

2

−

√

3

2

2

XXXXXXXXXXX

=

±

√

2

−

√

3

4

XXXXXXXXXXX

=

±

√

2

−

√

3

2

Since  

5

π

12

<

π

2

XXXX

5

π

12

is in quadrant 1

XXXX

→

cos

(

5

π

12

)

is positive

XXXX

XXXX

XXXX

(the negative solution is extraneous)

answer in pic more explain