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Evaluate $\dfrac{\sin{3x}}{\sin{x}}$ $-$ $\dfrac{\cos{3x}}{\cos{x}}$ by Triple angle identities

The sine of triple angle and cosine of triple angle functions are involved in forming a triple angle trigonometric expression in this trigonometry problem. The involvement of triple angle trigonometric functions expresses that the triple angle trigonometric identities should be used to evaluate the given triple angle trigonometric expression.

Now, let’s learn how to find the value of the sine of three times $x$ divided by sine of angle $x$ minus cosine of three times $x$ divided by cosine of angle $x$ by using the triple angle trigonometric formulas.

Expand the Triple angle Trigonometric functions

According to the triple angle identities, the sine of triple angle function $\sin{3x}$ can be expanded as per the sine triple angle identity and the cosine of triple angle function $\cos{3x}$ can also be expanded as per cosine triple angle identity.

$(1).\,\,$ $\sin{3x}$ $\,=\,$ $3\sin{x}$ $-$ $4\sin^3{x}$

$(2).\,\,$ $\cos{3x}$ $\,=\,$ $4\cos^3{x}$ $-$ $3\cos{x}$

Now, the triple angle trigonometric functions $\sin{3x}$ and $\cos{3x}$ can be replaced by their expansions in the trigonometric expression.

$\implies$ $\dfrac{\sin{3x}}{\sin{x}}$ $-$ $\dfrac{\cos{3x}}{\cos{x}}$ $\,=\,$ $\dfrac{3\sin{x}-4\sin^3{x}}{\sin{x}}$ $-$ $\dfrac{4\cos^3{x}-3\cos{x}}{\cos{x}}$

Release expression from rational by simplification

The terms in the numerator of each term in the trigonometric expression can be divided by the corresponding denominator to start simplifying the trigonometric expression.

$=\,\,$ $\dfrac{3\sin{x}}{\sin{x}}$ $-$ $\dfrac{4\sin^3{x}}{\sin{x}}$ $-$ $\bigg(\dfrac{4\cos^3{x}}{\cos{x}}$ $-$ $\dfrac{3\cos{x}}{\cos{x}}\bigg)$

Now, continue the procedure of simplifying the trigonometric expression.

$=\,\,$ $\dfrac{3\sin{x}}{\sin{x}}$ $-$ $\dfrac{4\sin^3{x}}{\sin{x}}$ $-$ $\dfrac{4\cos^3{x}}{\cos{x}}$ $+$ $\dfrac{3\cos{x}}{\cos{x}}$

$=\,\,$ $\dfrac{3\cancel{\sin{x}}}{\cancel{\sin{x}}}$ $-$ $\dfrac{4\cancel{\sin^3{x}}}{\cancel{\sin{x}}}$ $-$ $\dfrac{4\cancel{\cos^3{x}}}{\cancel{\cos{x}}}$ $+$ $\dfrac{3\cancel{\cos{x}}}{\cancel{\cos{x}}}$

$=\,\,$ $3$ $-$ $4\sin^2{x}$ $-$ $4\cos^2{x}$ $+$ $3$

$=\,\,$ $3$ $+$ $3$ $-$ $4\sin^2{x}$ $-$ $4\cos^2{x}$

$=\,\,$ $6$ $-$ $4\sin^2{x}$ $-$ $4\cos^2{x}$

Find the value of Trigonometric expression

The negative four is a common factor in both second and third terms. So, take out the negative four common from them.

$=\,\,$ $6$ $-$ $4 \times \big(\sin^2{x}+\cos^2{x}\big)$

The sum of squares of sine and cosine functions at an angle is one as per the Pythagorean identity of sine and cosine functions.

$=\,\,$ $6$ $-$ $4 \times (1)$

Now, find the value of the arithmetic expression by simplification.

$=\,\,$ $6$ $-$ $4 \times 1$

$=\,\,$ $6-4$

$=\,\,$ $2$

Another method

Evaluate $\dfrac{\sin{3x}}{\sin{x}}$ $-$ $\dfrac{\cos{3x}}{\cos{x}}$

Learn how to find the value of sine of three times angle $x$ divided by sine of $x$ minus cosine of three times $x$ divided by cosine of angle $x$ without using the triple angle trigonometric identities.

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