Question
Answers

Solve: $\tan 3A=\sin 45{}^\circ \cos 45{}^\circ +\sin 30{}^\circ $

Answer
VerifiedVerified
149.7k+ views
Hint: Take RHS and simplify the equation. Equate it to LHS. The value you get in RHS should be converted into the tan of the angle. Then equating it will get you the answer.

Complete step-by-step answer:
The trigonometric functions (also called circular functions, angle functions, or trigonometric functions) are real functions that relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. They are among the simplest periodic functions, and as such are also widely used for studying periodic phenomena, through Fourier analysis.
The most widely used trigonometric functions are the sine, the cosine, and the tangent. Their reciprocals are respectively the cosecant, the secant, and the cotangent, which are less used in modern mathematics.
The cosine function, along with sine and tangent, is one of the three most common trigonometric functions. In any right triangle, the cosine of an angle is the length of the adjacent side (A) divided by the length of the hypotenuse (H). In a formula, it is written simply as '$\cos $'. $\cos $ function (or cosine function) in a triangle is the ratio of the adjacent side to that of the hypotenuse. The cosine function is one of the three main primary trigonometric functions and it is itself the complement of sine (co + sine).
The cosine graph or the cos graph is an up-down graph just like the sine graph. The only difference between the sine graph and the cos graph is that the sine graph starts from $0$ while the cos graph starts from $90{}^\circ $ (or $\dfrac{\pi }{2}$).
So here LHS$=$RHS.
Now taking RHS$=\sin 45{}^\circ \cos 45{}^\circ +\sin 30{}^\circ $, we know that $\sin 45{}^\circ =\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}$ and $\sin 30{}^\circ =\dfrac{1}{2}$.
So substituting the above values in RHS,
RHS$=\dfrac{1}{\sqrt{2}}\times \dfrac{1}{\sqrt{2}}+\dfrac{1}{2}$
RHS$=\dfrac{1}{2}+\dfrac{1}{2}=1$
So we know,
 LHS$=$RHS and so, $\tan 3A=1$.
We know,
$\tan 45{}^\circ =1$
$\tan 3A=1=\tan 45{}^\circ $
So we get,
$3A=45{}^\circ $
We get,
$A=15{}^\circ $
So the value of $A$is $15{}^\circ $.

Note: Read the question carefully. Also, you must know the properties of trigonometric identities. While simplifying, do not miss any term. Also, while equating do not make mistakes. You must be familiar with the angles.