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
Answer : The exact value of [tex]\sin (\frac{5\pi}{12})=\frac{\sqrt{(2+\sqrt{3})}}{2}[/tex]
Step-by-step explanation :
As we are given that: [tex]\sin (\frac{5\pi}{12})[/tex]
Using a half-angle identity:
[tex]\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos\theta}{2}}[/tex]
[tex]\sin\left(\frac{\theta}{2}\right)=\sin \frac{(\frac{5\pi}{6})}{2}[/tex]
Thus, [tex]\theta=\frac{5\pi}{6}[/tex]
Now using half-angle identity, we get:
[tex]\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos\theta}{2}}[/tex]
[tex]\sin\left(\frac{(\frac{5\pi}{6})}{2}\right)=\sqrt{\frac{1-\cos (\frac{5\pi}{6})}{2}}[/tex]
[tex]\sin\left(\frac{(\frac{5\pi}{6})}{2}\right)=\sqrt{\frac{1-\cos (\pi -\frac{\pi}{6})}{2}}[/tex]
[tex]\sin\left(\frac{(\frac{5\pi}{6})}{2}\right)=\sqrt{\frac{1+\cos (\frac{\pi}{6})}{2}}[/tex]
As we know that,
[tex]\cos \frac{\pi}{6}=\cos 30^o=\frac{\sqrt{3}}{2}[/tex]
Now put the value of [tex]\cos \frac{\pi}{6}[/tex], we get:
[tex]\sin (\frac{5\pi}{12})=\sqrt{\frac{1+(\frac{\sqrt{3}}{2})}{2}}[/tex]
[tex]\sin (\frac{5\pi}{12})=\sqrt{\frac{(\frac{2+\sqrt{3}}{2})}{2}}[/tex]
[tex]\sin (\frac{5\pi}{12})=\sqrt{(\frac{2+\sqrt{3}}{4})}[/tex]
[tex]\sin (\frac{5\pi}{12})=\frac{\sqrt{(2+\sqrt{3})}}{2}[/tex]
Thus, the exact value of [tex]\sin (\frac{5\pi}{12})=\frac{\sqrt{(2+\sqrt{3})}}{2}[/tex]
