$3$ times sine of $3$ times angle $x$ is added to $2$ times the cosine of sum of $2$ times angle $x$ and $5$ degrees and it is divided by the sine of subtraction of $10$ degrees from $2$ times angle $x$, subtracted from $2$ times cosine of $3$ times angle $x$. The value of this rational function in terms of trigonometric function has to calculate when the angle of $x$ is equal to $20$ degrees.

$\dfrac{3\sin{3x}+2\cos{(2x+5^\circ)}}{2\cos{3x}-\sin{(2x-10^\circ)}}$

Now, let us learn how to evaluate this trigonometric rational function by substituting the value of angle $x$.

Now, substitute $x \,=\, 20^\circ$ in the rational expression in terms of trigonometric functions sine and cosine.

$=\,\,\,$ $\dfrac{3\sin{(3 \times 20^\circ)}+2\cos{(2 \times 20^\circ+5^\circ)}}{2\cos{(3 \times 20^\circ)}-\sin{(2 \times 20^\circ-10^\circ)}}$

It is time to simplify the expressions, which represent the angles inside each trigonometric function.

$=\,\,\,$ $\dfrac{3\sin{(60^\circ)}+2\cos{(40^\circ+5^\circ)}}{2\cos{(60^\circ)}-\sin{(40^\circ-10^\circ)}}$

$=\,\,\,$ $\dfrac{3\sin{(60^\circ)}+2\cos{(45^\circ)}}{2\cos{(60^\circ)}-\sin{(30^\circ)}}$

$=\,\,\,$ $\dfrac{3 \times \sin{(60^\circ)}+2 \times \cos{(45^\circ)}}{2 \times \cos{(60^\circ)}-\sin{(30^\circ)}}$

In the trigonometric rational function, the sine and cosine functions are appearing with three different angles. According to the trigonometry, substitute the exact values of sine of 60 degrees, cos of 45 degrees, cos of 60 degrees and sin of 30 degrees in fraction form.

$=\,\,\,$ $\dfrac{3 \times \dfrac{\sqrt{3}}{2}+2 \times \dfrac{1}{\sqrt{2}}}{2 \times \dfrac{1}{2}-\dfrac{1}{2}}$

The trigonometric expression becomes an arithmetic expression. So, it is very easier to find its value by simplification. The factors in fraction form in each term of both numerator and denominator can be calculated by the multiplication of fractions.

$=\,\,\,$ $\dfrac{\dfrac{3 \times \sqrt{3}}{2}+\dfrac{2 \times 1}{\sqrt{2}}}{\dfrac{2 \times 1}{2}-\dfrac{1}{2}}$

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

$=\,\,\,$ $\dfrac{\dfrac{3\sqrt{3}}{2}+\dfrac{2}{\sqrt{2}}}{\dfrac{\cancel{2}}{\cancel{2}}-\dfrac{1}{2}}$

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

Look at the denominator, it expresses the subtraction of fractions. So, find the difference of them as per the subtraction method of fractions.

$=\,\,\,$ $\dfrac{\dfrac{3\sqrt{3}}{2}+\dfrac{2}{\sqrt{2}}}{\dfrac{1 \times 2-1}{2}}$

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

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

The numerator and denominator both are in fraction form. So, the quotient of them can be calculated as per the division of the fractions.

$=\,\,\,$ $\bigg(\dfrac{3\sqrt{3}}{2}+\dfrac{2}{\sqrt{2}}\bigg) \times \dfrac{2}{1}$

$=\,\,\,$ $\bigg(\dfrac{3\sqrt{3}}{2}+\dfrac{2}{\sqrt{2}}\bigg) \times 2$

The above expression can be written in the following form as per the commutative property of multiplication.

$=\,\,\,$ $2 \times \bigg(\dfrac{3\sqrt{3}}{2}+\dfrac{2}{\sqrt{2}}\bigg)$

The factor 2 can be distributed over addition as per the distributive property.

$=\,\,\,$ $2 \times \dfrac{3\sqrt{3}}{2}+2 \times \dfrac{2}{\sqrt{2}}$

Now, let us simplify this mathematical expression to find the value of this expression. It can be done by using the multiplication of the fractions.

$=\,\,\,$ $\dfrac{2 \times 3\sqrt{3}}{2}+\dfrac{2 \times 2}{\sqrt{2}}$

For our convenience, the number $2$ can be written as the square of the square root of $2$.

$=\,\,\,$ $\dfrac{\cancel{2} \times 3\sqrt{3}}{\cancel{2}}+\dfrac{2 \times 2}{\sqrt{2}}$

$=\,\,\,$ $3\sqrt{3}+\dfrac{2 \times 2}{\sqrt{2}}$

$=\,\,\,$ $3\sqrt{3}+\dfrac{2 \times \big(\sqrt{2}\big)^2}{\sqrt{2}}$

$=\,\,\,$ $3\sqrt{3}+\dfrac{2 \times \sqrt{2} \times \sqrt{2}}{\sqrt{2}}$

$=\,\,\,$ $3\sqrt{3}+\dfrac{2 \times \sqrt{2} \times \cancel{\sqrt{2}}}{\cancel{\sqrt{2}}}$

$=\,\,\,$ $3\sqrt{3}+2 \times \sqrt{2}$

$=\,\,\,$ $3\sqrt{3}+2\sqrt{2}$

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