Values of sin 75, cos 75, tan 75 | sin 75 cos 75

In this post, we will find the value of sin 75 degree, cos 75 degree, and tan 75 degree. They are very important in Trigonometry. Using their values, we are able to find the trigonometric values of many angles.

We will now find the value of sine 75 degree.

Value of sin 75

Answer: The value of sin 75 is (√3+1)/2√2.

Explanation:

Let us find the value of $\sin 75$ using the formula of the compound angles of sine functions. The following formula is the key to find sin 75 degree:

sin(A+B) = sin A cos B + cos A sin B

Note that we can write sin 75 as follows:

$\sin 75 = \sin(45 +30)$

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

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

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

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

Thus, the value of sin 75 is (√3+1)/2√2.

Next, we will find the value of cosine 75.

Value of cos 75

The value of $\cos 75$ can be obtained using the value of $\sin 75$. Here we will use the following formula:

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

Put $x=75.$

So we get that

$\sin^2 75 + \cos^2 75=1$

$\Rightarrow \left( \dfrac{√3+1}{2√2} \right)^2 + \cos^2 75=1$ as sin 75 = (√3+1)/2√2

$\Rightarrow  \cos^2 75=1-\left( \dfrac{√3+1}{2√2} \right)^2$

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

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

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

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

Taking square root on both sides, we get that

$\cos 75 = \dfrac{\sqrt{3}-1}{2\sqrt{2}}$

So we have obtained the value of cos 75 which is (√3-1)/2√2.

Also Read:

Sinx=0, cosx=0, tanx=0 General Solution

Values of sin 15, cos 15, tan 15

sin 75 cos 75

Question: Find the value of sin 75 cos 75.

Answer:

Using the values of $\sin 75$ and $\cos 75$ we can compute the value of the product $\sin 75 \cos 75.$

As we know from above that sin 75 = (√3+1)/2√2 and cos 75 = (√3-1)/2√2, so we get that

sin 75 cos 75 = (√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

So the value of sin75cos75 is equal to 1/4.

Value of tan 75

(Method 1 of finding tan 75:) At first, we will evaluate the value of $\tan 75$ using the values of $\sin 75$ and $\cos 75.$

From above, we have sin 75 = (√3+1)/2√2 and cos 75 = (√3-1)/2√2.

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

$\tan 75 =\dfrac{\sin 75}{\cos 75}$

$\therefore \tan 75 = \dfrac{\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}-1}{2\sqrt{2}}}$

$= \dfrac{\sqrt{3}+1}{\sqrt{3}-1}$

$=\dfrac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)}$ 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 75 is 2+√3.

(Method 2 of finding tan 75:) Next, we will find the value of $\tan 75$ 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

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

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

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

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

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

So 2+√3 is the value of tan 75.

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