# Values of sin 15, cos 15, tan 15 | sin 15 cos 15

The values of sin 15°, cos 15°, and tan 15° are very important in the theory of Trigonometry. We will find their values in this post.

Let us now find the value of sin 15 degree.

## Value of sin 15

We will evaluate the value of $\sin 15$ using the formula of the compound angles of sine functions. We will use the following formula:

sin(A-B) = sin A cos B – cos A sin B

Note that $\sin 15 = \sin(45 -30)$

$= \sin 45 \cdot \cos30 – \cos 45 \cdot \sin 30$

$= \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} – \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2}$

$= \dfrac{\sqrt{3}}{2\sqrt{2}} – \dfrac{1}{2\sqrt{2}}$

$= \dfrac{\sqrt{3}-1}{2\sqrt{2}}$

Thus the value of sin 15 is (√3−1)/2√2.

We will now find the value of cos 15.

## Value of cos 15

The value of $\cos 15$ can be obtained using the value of $\sin 15$. Here we will use the following formula:

$\sin^2 x + \cos^2 x=1$

Put $x=15.$ So we get that

$\sin^2 15 + \cos^2 15=1$

$\Rightarrow \left(\dfrac{(\sqrt{3}−1)}{2\sqrt{2}} \right)^2 + \cos^2 15=1$ as sin 15 = (√3−1)/2√2

$\Rightarrow \cos^2 15=1-\left( \dfrac{\sqrt{3}−1}{2\sqrt{2}} \right)^2$

$=\dfrac{8-3+2\sqrt{3}-1}{(2\sqrt{2})^2}$

$=\dfrac{4+2\sqrt{3}}{(2\sqrt{2})^2}$

$=\dfrac{3+2\sqrt{3}+1}{(2\sqrt{2})^2}$

$=\dfrac{(\sqrt{3}+1)^2}{(2\sqrt{2})^2}$

Taking square root on both sides, we get that

$\cos 15 = \dfrac{\sqrt{3}+1}{2\sqrt{2}}$

So we have obtained the value of cos 15 which is (√3+1)/2√2.

Sinx=0 General Solution

## sin 15 cos 15

Question: Find the value of sin 15 cos 15.

Using the values of $\sin 15$ and $\cos 15$ we can compute the value of the product $\sin 15 \cos 15.$

As we know from above that sin 15 = (√3-1)/2√2 and cos 15 = (√3+1)/2√2, so we get that

sin 15 cos 15 = (√3-1)/2√2  × (√3+1)/2√2

= [(√3-1)(√3-1)] / (2√2)2

= [(√3)2 – 12] / 8 by the formula of a2 -b2

= (3-1)/8

= 2/8 = 1/4

## Value of tan 15

(Method 1 of finding tan 15:) At first, we will find the value of $\tan 15$ using the values of $\sin 15$ and $\cos 15.$

From above, we have sin 15 = (√3-1)/2√2 and cos 15 = (√3+1)/2√2.

As $\tan x =\dfrac{\sin x}{\cos x}$ we obtain that

$\tan 15 =\dfrac{\sin 15}{\cos 15}$

$\therefore \tan 15 = \dfrac{\dfrac{\sqrt{3}-1}{2\sqrt{2}}}{\dfrac{\sqrt{3}+1}{2\sqrt{2}}}$

$= \dfrac{\sqrt{3}-1}{\sqrt{3}+1}$

$=\dfrac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)}$ rationalizing the denominator

$=\dfrac{3-2\sqrt{3}+1}{(\sqrt{3})^2-1^2}$

$=\dfrac{4-2\sqrt{3}}{3-1}$

$=\dfrac{2(2-\sqrt{3})}{2}$

$=2-\sqrt{3}$

Thus the value of tan 15 is 2-√3.

(Method 2 of finding tan 15:) Next, we will find the value of $\tan 15$ using the difference formula of two angles for tangent. The formula is given below.

$\tan (A-B)=\dfrac{\tan A -\tan B}{1+\tan A \tan B}$

Put $A=45$ and $B=30$. So we have

$\tan(45-30)$ $=\dfrac{\tan 45 -\tan 30}{1+\tan 45 \tan 30}$

$=\dfrac{1 -\dfrac{1}{\sqrt{3}}}{1+1 \cdot \dfrac{1}{\sqrt{3}}}$

$=\dfrac{\dfrac{\sqrt{3}-1}{\sqrt{3}}}{\dfrac{\sqrt{3}+1}{\sqrt{3}}}$

$=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$

$=2-\sqrt{3}$ rationalizing the denominator as above.

So 2-√3 is the value of tan 15.

## FAQs

Q1: What is the value of sin15?

Answer: The value of sin 15 is (√3−1)/2√2.

Q2: What is the value of cos15?

Answer: The value of cos 15 is (√3+1)/2√2.

Q3: What is the value of tan15?

Answer: The value of tan15 is 2-√3.