
How do you find \[\sin 2a\] using sum identity?
Answer
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Hint: Here, we have to find \[\sin 2a\] using sum identity. For this, we first need to know how to write \[2a\] as a sum of two angles. We will write \[2a\] as \[\left( {a + a} \right)\]. Then we will use the angle sum identity of the sine function which is given by \[\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y\]. Here, we will put both \[x\] and \[y\] as \[a\]. Then we will simplify it to find the result.
Complete step by step answer:
We have to find \[\sin 2a\] using the sum identity.
We need to write \[2a\] as \[\left( {a + a} \right)\] in \[\sin 2a\].
On writing \[2a\] as \[\left( {a + a} \right)\], we get
\[ \Rightarrow \sin 2a = \sin \left( {a + a} \right)\]
We know that the angle sum identity of the sine function is given by \[\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y\].
Here, we have both \[x\] and \[y\] as \[a\].
Using the angle sum identity, we can write \[\sin 2a\] as
\[ \Rightarrow \sin 2a = \sin a\cos a + \cos a\sin a\]
On simplification, we get
\[ \therefore \sin 2a = 2\sin a\cos a\].
This is the required answer.
Note:
> Trigonometric formulas are considered only for right angled triangles. We have three sides namely hypotenuse, perpendicular and base in a right-angled triangle. Hypotenuse is the longest side, side opposite to the angle is perpendicular and base is the side where both hypotenuse and perpendicular rests. Basically, there are six ratios for finding the elements in trigonometry.
> Here, if we have cosine then we will use the formula \[\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y\]. Similarly, we have \[\sin \left( {x - y} \right) = \sin x\cos y - \cos x\sin y\] and \[\cos \left( {x - y} \right) = \cos x\cos y + \sin x\sin y\]. Also, note that when we consider all four quadrants, we get to know that all trigonometric functions are positive in the first quadrant, only sine and cosec is positive in second quadrant, only tan and cot are positive in the third quadrant and only secant and cosine in positive in the fourth quadrant.
Complete step by step answer:
We have to find \[\sin 2a\] using the sum identity.
We need to write \[2a\] as \[\left( {a + a} \right)\] in \[\sin 2a\].
On writing \[2a\] as \[\left( {a + a} \right)\], we get
\[ \Rightarrow \sin 2a = \sin \left( {a + a} \right)\]
We know that the angle sum identity of the sine function is given by \[\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y\].
Here, we have both \[x\] and \[y\] as \[a\].
Using the angle sum identity, we can write \[\sin 2a\] as
\[ \Rightarrow \sin 2a = \sin a\cos a + \cos a\sin a\]
On simplification, we get
\[ \therefore \sin 2a = 2\sin a\cos a\].
This is the required answer.
Note:
> Trigonometric formulas are considered only for right angled triangles. We have three sides namely hypotenuse, perpendicular and base in a right-angled triangle. Hypotenuse is the longest side, side opposite to the angle is perpendicular and base is the side where both hypotenuse and perpendicular rests. Basically, there are six ratios for finding the elements in trigonometry.
> Here, if we have cosine then we will use the formula \[\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y\]. Similarly, we have \[\sin \left( {x - y} \right) = \sin x\cos y - \cos x\sin y\] and \[\cos \left( {x - y} \right) = \cos x\cos y + \sin x\sin y\]. Also, note that when we consider all four quadrants, we get to know that all trigonometric functions are positive in the first quadrant, only sine and cosec is positive in second quadrant, only tan and cot are positive in the third quadrant and only secant and cosine in positive in the fourth quadrant.
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