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# If $\sin \alpha =\dfrac{336}{625}$ and $450{}^\circ < \alpha < 540{}^\circ$, then find the value of $\sin \dfrac{\alpha }{4}$.

Last updated date: 24th Jul 2024
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Answer
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Hint: This question belongs to the topic of trigonometry. In this question, first we will convert the function of sin in terms of cos function. Using the half angle formula of cos function two times, we will solve this question. Half angle formula is: $\cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x$. We will use the condition $450{}^\circ < \alpha < 540{}^\circ$ to get the batter solution of this question.

Complete step by step answer:
Let us solve this question.
In this question, it is given that $\sin \alpha =\dfrac{336}{625}$ and we have to find the value of $\sin \dfrac{\alpha }{4}$ in which we have given condition that $450{}^\circ < \alpha < 540{}^\circ$.
The term $\sin \alpha =\dfrac{336}{625}$ is in sin form. Let us make this in cos form using the formula: ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$
We can write
${{\left( \dfrac{336}{625} \right)}^{2}}+{{\cos }^{2}}\alpha =1$
The above equation can also be written as
$\Rightarrow {{\cos }^{2}}\alpha =1-{{\left( \dfrac{336}{625} \right)}^{2}}=1-{{\dfrac{\left( 336 \right)}{{{\left( 625 \right)}^{2}}}}^{2}}$
The above equation can also be written as
$\Rightarrow {{\cos }^{2}}\alpha ={{\dfrac{{{\left( 625 \right)}^{2}}-\left( 336 \right)}{{{\left( 625 \right)}^{2}}}}^{2}}$
Using the formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$, we can write the above equation as
$\Rightarrow {{\cos }^{2}}\alpha =\dfrac{\left( 625-336 \right)\left( 625+336 \right)}{{{\left( 625 \right)}^{2}}}$
The above equation can also be written as
$\Rightarrow {{\cos }^{2}}\alpha =\dfrac{\left( 289 \right)\left( 961 \right)}{{{\left( 625 \right)}^{2}}}$
As we know that square of 17 is 289 and the square of 19 is 961. Then, we can write the above equation as
$\Rightarrow {{\cos }^{2}}\alpha =\dfrac{{{\left( 17 \right)}^{2}}{{\left( 31 \right)}^{2}}}{{{\left( 625 \right)}^{2}}}$
Now, squaring root in both the side of equation, we get
$\Rightarrow \cos \alpha =\sqrt{\dfrac{{{\left( 17 \right)}^{2}}{{\left( 31 \right)}^{2}}}{{{\left( 625 \right)}^{2}}}}=\pm \dfrac{\left( 17 \right)\left( 31 \right)}{\left( 625 \right)}$
The above equation can also be written as
$\Rightarrow \cos \alpha =\pm \dfrac{527}{625}$
The question has given a condition that $450{}^\circ < \alpha < 540{}^\circ$. As we know that the period of sin and cos function is $360{}^\circ$. So, after $360{}^\circ$, it will repeat the values. Hence, we can write $450{}^\circ < \alpha < 540{}^\circ$ as
$360{}^\circ +90{}^\circ < \alpha < 360{}^\circ +180{}^\circ$
As the functions repeat after $360{}^\circ$, so we can write
$90{}^\circ < \alpha < 180{}^\circ$
Hence, we get the range of $\alpha$.
Now, we know that the value of cos is negative in this range and it cannot be positive.
So, from the equation $\cos \alpha =\pm \dfrac{527}{625}$, we can write
$\Rightarrow \cos \alpha =-\dfrac{527}{625}$
Now, using the formula $\cos 2x=2{{\cos }^{2}}x-1$, we can write $\cos \alpha$ as $\cos \alpha =2{{\cos }^{2}}\dfrac{\alpha }{2}-1$. So, the above equation can also be written as
$\Rightarrow 2{{\cos }^{2}}\dfrac{\alpha }{2}-1=-\dfrac{527}{625}$
The above equation can also be written as
$\Rightarrow 2{{\cos }^{2}}\dfrac{\alpha }{2}=1-\dfrac{527}{625}=\dfrac{625-527}{625}$
The above equation can also be written as
$\Rightarrow 2{{\cos }^{2}}\dfrac{\alpha }{2}=\dfrac{98}{625}$
By dividing 2 on both sides of the above equation, we can write
$\Rightarrow {{\cos }^{2}}\dfrac{\alpha }{2}=\dfrac{49}{625}$
Now, squaring root both sides, we get
$\Rightarrow \cos \dfrac{\alpha }{2}=\pm \sqrt{\dfrac{49}{625}}$
As we know that the square root of 49 is 7 and the square of 625 is 25.
So, we can write the above equation as
$\Rightarrow \cos \dfrac{\alpha }{2}=\pm \dfrac{7}{25}$
According to the condition $450{}^\circ < \alpha < 540{}^\circ$, we can write it as
$\dfrac{450{}^\circ }{2} < \dfrac{\alpha }{2} < \dfrac{540{}^\circ }{2}$
The above can also be written as
$225{}^\circ < \dfrac{\alpha }{2} < 270{}^\circ$
As we can that the value of $\dfrac{\alpha }{2}$ is in the third quadrant (third quadrant has the range of 180 degrees to 270 degrees)
And, we know that the value of the cos function is negative (not positive) in the third quadrant.
So, we can write the equation $\cos \dfrac{\alpha }{2}=\pm \dfrac{7}{25}$ as
$\Rightarrow \cos \dfrac{\alpha }{2}=-\dfrac{7}{25}$
Let us convert this in sin form using the formula $\cos 2x=1-2{{\sin }^{2}}x$, we can write $\cos \dfrac{\alpha }{2}=1-2{{\sin }^{2}}\dfrac{\alpha }{4}$.
The above equation can also be written as
$\Rightarrow 1-2{{\sin }^{2}}\dfrac{\alpha }{4}=-\dfrac{7}{25}$
We can write the above equation as
$\Rightarrow 1+\dfrac{7}{25}=2{{\sin }^{2}}\dfrac{\alpha }{4}$
We can write the above equation as
$\Rightarrow 2{{\sin }^{2}}\dfrac{\alpha }{4}=\dfrac{25+7}{25}=\dfrac{32}{25}$
The above equation can also be written as
$\Rightarrow {{\sin }^{2}}\dfrac{\alpha }{4}=\dfrac{16}{25}$
After squaring root both side, we get
$\Rightarrow \sin \dfrac{\alpha }{4}=\pm \sqrt{\dfrac{16}{25}}$
As we know that 16 is a square of 4 and 25 is a square of 5. So, we can write
$\Rightarrow \sin \dfrac{\alpha }{4}=\pm \dfrac{4}{5}$
We have given a condition that $450{}^\circ < \alpha < 540{}^\circ$, we can write this as
$112.5{}^\circ < \dfrac{\alpha }{4} < 135{}^\circ$, this range lies in the second quadrant. The value of sin in the second quadrant is positive. So, we can write
$\Rightarrow \sin \dfrac{\alpha }{4}=\dfrac{4}{5}$

Now, we have found the value of $\sin \dfrac{\alpha }{4}$ which is $\dfrac{4}{5}$.

Note: We should have a very deep knowledge in the topic of trigonometry to solve this type of question.
Remember the formulas:
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
$\cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x$
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
And always remember that
1) In the first quadrant (from 0 degree to 90 degrees), the value of sin and cos is positive
2) In the second quadrant (from 90 degrees to 180 degrees), the value of sin is positive and the value of cos is negative
3) In the third quadrant (from 180 degrees to 270 degrees), the value of sin and cos is negative
4) In the fourth quadrant (from 270 degrees to 360 degrees), the value of cos is positive and the value of sin is negative.