# Prove $\sin 3x{{\sin }^{3}}x+\cos 3x{{\cos }^{3}}x={{\cos }^{3}}2x$.

Last updated date: 25th Mar 2023

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Hint: We have to prove $\sin 3x{{\sin }^{3}}x+\cos 3x{{\cos }^{3}}x={{\cos }^{3}}2x$. For that we have to take LHS. So take LHS and use different trigonometric identities. Also, Use different algebraic identities. Try it and you will get 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 $\dfrac{\pi }{2}$).

Now we have to prove LHS$=$RHS.

So now taking LHS$=\sin 3x{{\sin }^{3}}x+\cos 3x{{\cos }^{3}}x$, we know that,

$\begin{align}

& \sin 3x=3\sin x-4{{\sin }^{3}}x=\sin x(3-4{{\sin }^{2}}x) \\

& \cos 3x=4{{\cos }^{3}}x-3\cos x=\cos x(4{{\cos }^{2}}x-3) \\

\end{align}$

\[\begin{align}

& =\sin x(3-4{{\sin }^{2}}x){{\sin }^{3}}x+\cos x(4{{\cos }^{2}}x-3){{\cos }^{3}}x \\

& =(3-4{{\sin }^{2}}x){{\sin }^{4}}x+(4{{\cos }^{2}}x-3){{\cos }^{4}}x \\

& =3{{\sin }^{4}}x-4{{\sin }^{6}}x+4{{\cos }^{6}}x-3{{\cos }^{4}}x \\

& =3{{\sin }^{4}}x-3{{\cos }^{4}}x-4{{\sin }^{6}}x+4{{\cos }^{6}}x \\

& =3({{\sin }^{4}}x-{{\cos }^{4}}x)-4({{\sin }^{6}}x-{{\cos }^{6}}x) \\

\end{align}\]

Now we know that,

${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$and ${{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})$

Also, \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]

So applying above identity we get,

\[\begin{align}

& =3({{\sin }^{2}}x-{{\cos }^{2}}x)({{\sin }^{2}}x+{{\cos }^{2}}x)-4({{\sin }^{2}}x-{{\cos }^{2}}x)({{\sin }^{4}}x+{{\cos }^{4}}x+{{\sin }^{2}}x{{\cos }^{2}}x) \\

& =3({{\sin }^{2}}x-{{\cos }^{2}}x)-4({{\sin }^{2}}x-{{\cos }^{2}}x)({{\sin }^{4}}x+{{\cos }^{4}}x+{{\sin }^{2}}x{{\cos }^{2}}x) \\

& =3({{\sin }^{2}}x-{{\cos }^{2}}x)-4({{\sin }^{2}}x-{{\cos }^{2}}x)({{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{^{2}}}-2{{\sin }^{2}}x{{\cos }^{2}}x+{{\sin }^{2}}x{{\cos }^{2}}x) \\

& =3({{\sin }^{2}}x-{{\cos }^{2}}x)-4({{\sin }^{2}}x-{{\cos }^{2}}x)(1-{{\sin }^{2}}x{{\cos }^{2}}x) \\

\end{align}\]

\[\begin{align}

& =-3(-{{\sin }^{2}}x+{{\cos }^{2}}x)+4({{\cos }^{2}}x-{{\sin }^{2}}x)(1-{{\sin }^{2}}x{{\cos }^{2}}x) \\

& =4({{\cos }^{2}}x-{{\sin }^{2}}x)(1-{{\sin }^{2}}x{{\cos }^{2}}x)-3({{\cos }^{2}}x-{{\sin }^{2}}x) \\

\end{align}\]

Now taking \[{{\cos }^{2}}x-{{\sin }^{2}}x\]common we get,

Also, \[({{\cos }^{2}}x-{{\sin }^{2}}x)=\cos 2x\]

\[\begin{align}

& =({{\cos }^{2}}x-{{\sin }^{2}}x)\left( 4(1-{{\sin }^{2}}x{{\cos }^{2}}x)-3 \right) \\

& =\cos 2x\left( 4-4{{\sin }^{2}}x{{\cos }^{2}}x-3 \right) \\

& =\cos 2x\left( 4-4(1-{{\cos }^{2}}x){{\cos }^{2}}x-3 \right) \\

& =\cos 2x\left( 1-4{{\cos }^{2}}x+4{{\cos }^{4}}x \right) \\

& =\cos 2x\left( 1-2\times 2{{\cos }^{2}}x+{{(2{{\cos }^{2}}x)}^{2}} \right) \\

& =\cos 2x{{\left( 2{{\cos }^{2}}x-1 \right)}^{2}} \\

\end{align}\]

We know that\[2{{\cos }^{2}}x-1=\cos 2x\],

\[\begin{align}

& =\cos 2x{{\left( \cos 2x \right)}^{2}} \\

& ={{\cos }^{3}}2x \\

\end{align}\]

\[={{\cos }^{3}}2x=\]RHS

So $\sin 3x{{\sin }^{3}}x+\cos 3x{{\cos }^{3}}x={{\cos }^{3}}2x$ is proved.

Hence proved.

Note: Read the question in a careful manner. Also, make sure that you know all the identities. You should be familiar with the trigonometric formulae because mostly they are used for conversion. Also, there might be silly mistakes, so kindly avoid the silly mistake.

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 $\dfrac{\pi }{2}$).

Now we have to prove LHS$=$RHS.

So now taking LHS$=\sin 3x{{\sin }^{3}}x+\cos 3x{{\cos }^{3}}x$, we know that,

$\begin{align}

& \sin 3x=3\sin x-4{{\sin }^{3}}x=\sin x(3-4{{\sin }^{2}}x) \\

& \cos 3x=4{{\cos }^{3}}x-3\cos x=\cos x(4{{\cos }^{2}}x-3) \\

\end{align}$

\[\begin{align}

& =\sin x(3-4{{\sin }^{2}}x){{\sin }^{3}}x+\cos x(4{{\cos }^{2}}x-3){{\cos }^{3}}x \\

& =(3-4{{\sin }^{2}}x){{\sin }^{4}}x+(4{{\cos }^{2}}x-3){{\cos }^{4}}x \\

& =3{{\sin }^{4}}x-4{{\sin }^{6}}x+4{{\cos }^{6}}x-3{{\cos }^{4}}x \\

& =3{{\sin }^{4}}x-3{{\cos }^{4}}x-4{{\sin }^{6}}x+4{{\cos }^{6}}x \\

& =3({{\sin }^{4}}x-{{\cos }^{4}}x)-4({{\sin }^{6}}x-{{\cos }^{6}}x) \\

\end{align}\]

Now we know that,

${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$and ${{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})$

Also, \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]

So applying above identity we get,

\[\begin{align}

& =3({{\sin }^{2}}x-{{\cos }^{2}}x)({{\sin }^{2}}x+{{\cos }^{2}}x)-4({{\sin }^{2}}x-{{\cos }^{2}}x)({{\sin }^{4}}x+{{\cos }^{4}}x+{{\sin }^{2}}x{{\cos }^{2}}x) \\

& =3({{\sin }^{2}}x-{{\cos }^{2}}x)-4({{\sin }^{2}}x-{{\cos }^{2}}x)({{\sin }^{4}}x+{{\cos }^{4}}x+{{\sin }^{2}}x{{\cos }^{2}}x) \\

& =3({{\sin }^{2}}x-{{\cos }^{2}}x)-4({{\sin }^{2}}x-{{\cos }^{2}}x)({{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{^{2}}}-2{{\sin }^{2}}x{{\cos }^{2}}x+{{\sin }^{2}}x{{\cos }^{2}}x) \\

& =3({{\sin }^{2}}x-{{\cos }^{2}}x)-4({{\sin }^{2}}x-{{\cos }^{2}}x)(1-{{\sin }^{2}}x{{\cos }^{2}}x) \\

\end{align}\]

\[\begin{align}

& =-3(-{{\sin }^{2}}x+{{\cos }^{2}}x)+4({{\cos }^{2}}x-{{\sin }^{2}}x)(1-{{\sin }^{2}}x{{\cos }^{2}}x) \\

& =4({{\cos }^{2}}x-{{\sin }^{2}}x)(1-{{\sin }^{2}}x{{\cos }^{2}}x)-3({{\cos }^{2}}x-{{\sin }^{2}}x) \\

\end{align}\]

Now taking \[{{\cos }^{2}}x-{{\sin }^{2}}x\]common we get,

Also, \[({{\cos }^{2}}x-{{\sin }^{2}}x)=\cos 2x\]

\[\begin{align}

& =({{\cos }^{2}}x-{{\sin }^{2}}x)\left( 4(1-{{\sin }^{2}}x{{\cos }^{2}}x)-3 \right) \\

& =\cos 2x\left( 4-4{{\sin }^{2}}x{{\cos }^{2}}x-3 \right) \\

& =\cos 2x\left( 4-4(1-{{\cos }^{2}}x){{\cos }^{2}}x-3 \right) \\

& =\cos 2x\left( 1-4{{\cos }^{2}}x+4{{\cos }^{4}}x \right) \\

& =\cos 2x\left( 1-2\times 2{{\cos }^{2}}x+{{(2{{\cos }^{2}}x)}^{2}} \right) \\

& =\cos 2x{{\left( 2{{\cos }^{2}}x-1 \right)}^{2}} \\

\end{align}\]

We know that\[2{{\cos }^{2}}x-1=\cos 2x\],

\[\begin{align}

& =\cos 2x{{\left( \cos 2x \right)}^{2}} \\

& ={{\cos }^{3}}2x \\

\end{align}\]

\[={{\cos }^{3}}2x=\]RHS

So $\sin 3x{{\sin }^{3}}x+\cos 3x{{\cos }^{3}}x={{\cos }^{3}}2x$ is proved.

Hence proved.

Note: Read the question in a careful manner. Also, make sure that you know all the identities. You should be familiar with the trigonometric formulae because mostly they are used for conversion. Also, there might be silly mistakes, so kindly avoid the silly mistake.

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