Answer

Verified

407.1k+ views

**Hint:**In the given problem we have to find the value of $\cos 15$ by using the angle of difference identity. There are various problems. Here we have to use the identity of $\cos \left( A-B \right)$ Simplify the answer step wise properly and also you need to know the proper value of sine and cosine of the given angle.

**Complete step by step solution:**Here we have, $\cos 15{}^\circ $

If we can find the two angles $A$ and $B$ where whose sum and difference is $15{}^\circ $ and also whose sine and cosine we know,

Therefore we have to use difference identity,

$\cos \left( A-B \right)=\cos A.\cos B+\sin A.\sin B$

We can write $\cos 15{}^\circ $ as,

$\cos 15{}^\circ =\cos \left( 60{}^\circ -45{}^\circ \right)$

Here, $\left( 60-45 \right)=15$ and we know the trigonometric function of $60$ and $45.$

Substitute it in the given identity then the equation becomes,

$\cos 15{}^\circ =\cos 60{}^\circ \cos 45{}^\circ +\sin 60{}^\circ \sin 45{}^\circ $

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

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

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

Hence,

The value of $\cos 15{}^\circ =\dfrac{1+\sqrt{3}}{2\sqrt{2}}$

**Additional Information:**

We have to use the sum and difference identities for sin, cosine and tangent. While solving any trigonometric problem you need to convert all $\sec ,\cos ,\csc $ and to $\sin $ and this helps us to solve it easily. Using identities then check all sums and differences and use identities to remove what you have to remove.

If you have any factored problem so the common factor can be canceled from the numerator and denominator while solving the problem.

You don't always have to use $\sin $ and $\cos $ always but it is easier to compare when both sides are composed of similar functions. If you are practicing let others read through your problem just to check whether they can give a new way. If we stare at an equation for a long time, then we start to miss any ideas that came in our mind.

**Note:**

Always check properly what is asking in question because if there is $\sin 15$ the students sometimes use $\cos $ identity there. We could use similar property for $\sin 15$ but only we have to change the identity there. We have to use $\sin \left( A-B \right)$ identity. We should use $45-30=15$ instead of $60-45=15$ for $\sin 15.$ Main point is that there is no other way for finding this type of solution.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE