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Question: If \(\sin \theta =\dfrac{1}{4}\), \(\theta \) being in quadrant II, how do you find exact value of \...

If sin⁡θ=14\sin \theta =\dfrac{1}{4}, θ\theta being in quadrant II, how do you find exact value of sin⁡(θ−π3)\sin \left( \theta -\dfrac{\pi }{3} \right) ?

Explanation

Solution

We explain the function sin⁡θ=14\sin \theta =\dfrac{1}{4} and the quadrant value for the angle θ\theta . We express the identity functions of other ratio of cos with ratio of sin. It’s given that sin⁡θ=14\sin \theta =\dfrac{1}{4}. Thereafter we put the value to find the value of each of the remaining trigonometric functions. We also use the associative angle formula of sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B.

Complete step by step answer:
It’s given that sin⁡θ=14\sin \theta =\dfrac{1}{4}, θ\theta being in quadrant II. In that quadrant only sin⁡θ\sin \theta and csc⁡θ\csc \theta are positive whereas all the other ratios are negative.
We know the sum of the square law of (sin⁡x)2+(cos⁡x)2=1{{\left( \sin x \right)}^{2}}+{{\left( \cos x \right)}^{2}}=1.
We can put the value of sin⁡θ=14\sin \theta =\dfrac{1}{4} in the equation of (sin⁡x)2+(cos⁡x)2=1{{\left( \sin x \right)}^{2}}+{{\left( \cos x \right)}^{2}}=1.
Putting the value of sin⁡θ\sin \theta , we get (14)2+(cos⁡θ)2=1{{\left( \dfrac{1}{4} \right)}^{2}}+{{\left( \cos \theta \right)}^{2}}=1.
Now we perform the binary operations.
(14)2+(cos⁡θ)2=1 ⇒(cos⁡θ)2=1−116=1516 ⇒(cos⁡θ)=±154 {{\left( \dfrac{1}{4} \right)}^{2}}+{{\left( \cos \theta \right)}^{2}}=1 \\\ \Rightarrow {{\left( \cos \theta \right)}^{2}}=1-\dfrac{1}{16}=\dfrac{15}{16} \\\ \Rightarrow \left( \cos \theta \right)=\pm \dfrac{\sqrt{15}}{4} \\\
The value of cos⁡θ\cos \theta in quadrant II will be negative and that’s why (cos⁡θ)=−154\left( \cos \theta \right)=-\dfrac{\sqrt{15}}{4}.
Now we use the identity formula of sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B for sin⁡(θ−π3)\sin \left( \theta -\dfrac{\pi }{3} \right).
So, sin⁡(θ−π3)=sin⁡θcos⁡π3−cos⁡θsin⁡π3\sin \left( \theta -\dfrac{\pi }{3} \right)=\sin \theta \cos \dfrac{\pi }{3}-\cos \theta \sin \dfrac{\pi }{3}.
Putting the values, we get,
sin⁡(θ−π3)=14×12−(−154)×32 ⇒sin⁡(θ−π3)=18+358 ∴sin⁡(θ−π3)=1+358\sin \left( \theta -\dfrac{\pi }{3} \right)=\dfrac{1}{4}\times \dfrac{1}{2}-\left( -\dfrac{\sqrt{15}}{4} \right)\times \dfrac{\sqrt{3}}{2}\\\ \Rightarrow\sin \left( \theta -\dfrac{\pi }{3} \right) =\dfrac{1}{8}+\dfrac{3\sqrt{5}}{8}\\\ \therefore\sin \left( \theta -\dfrac{\pi }{3} \right) =\dfrac{1+3\sqrt{5}}{8}

Therefore, the exact value of sin⁡(θ−π3)\sin \left( \theta -\dfrac{\pi }{3} \right) is 1+358\dfrac{1+3\sqrt{5}}{8}.

Note: In addition to the reciprocal relationships of certain trigonometric functions, two other types of relationships exist. These relationships, known as trigonometric identities, include functional relationships and Pythagorean relationships.