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.

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sinx=1 Solution

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.

Also Read:

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Sinx=0 General Solution

sin 15 cos 15

Question: Find the value of sin 15 cos 15.

Answer:

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.

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