
If $A={{45}^{0}}$ Verify that
$\sin (2A)=2\sin A\cos A$
Answer
607.8k+ views
Hint: The Trigonometric ratios table helps to find the values of trigonometric standard angles such as ${{0}^{0}},{{30}^{0}},{{45}^{0}},{{60}^{0}}$ and ${{90}^{0}}$. It consists of trigonometric ratios – sine, cosine, tangent, cosecant, secant and cotangent. These ratios can be written in short as sin, cos, tan, cosec, sec and cot.
Complete step-by-step answer:
The value of the trigonometric ratios by using the trigonometric table is given below.
$\sin {{45}^{0}}=\dfrac{1}{\sqrt{2}},\cos {{45}^{0}}=\dfrac{1}{\sqrt{2}},\sin {{90}^{0}}=1$
Let us consider the right side of the given expression and put $A={{45}^{0}}$, we get
$2\sin A\cos A=2\sin {{45}^{0}}\cos {{45}^{0}}$
\[2\sin A\cos A=2\times \dfrac{1}{\sqrt{2}}\times \dfrac{1}{\sqrt{2}}\]
Multiplying the terms on the right side, we get
\[2\sin A\cos A=2\times \dfrac{1}{2}\]
Cancelling the term 2 on the right side, we get
\[2\sin A\cos A=1\]
From the trigonometric table, $1=\sin {{90}^{0}}$
$2\sin A\cos A=\sin {{90}^{0}}$
$2\sin A\cos A=\sin \left( 2\times {{45}^{0}} \right)$
Now put $A={{45}^{0}}$, we get
$2\sin A\cos A=\sin \left( 2A \right)$
Hence the given expression is verified
Note: Doubling the $\sin x$ will not give you the value of $\sin 2x$ . Nor will taking half of $\sin x$, give you $\sin \left( \dfrac{x}{2} \right)$ . We can develop the double angle formulas directly by using the addition formulas for sine, cosine and tangent.
Complete step-by-step answer:
The value of the trigonometric ratios by using the trigonometric table is given below.
$\sin {{45}^{0}}=\dfrac{1}{\sqrt{2}},\cos {{45}^{0}}=\dfrac{1}{\sqrt{2}},\sin {{90}^{0}}=1$
Let us consider the right side of the given expression and put $A={{45}^{0}}$, we get
$2\sin A\cos A=2\sin {{45}^{0}}\cos {{45}^{0}}$
\[2\sin A\cos A=2\times \dfrac{1}{\sqrt{2}}\times \dfrac{1}{\sqrt{2}}\]
Multiplying the terms on the right side, we get
\[2\sin A\cos A=2\times \dfrac{1}{2}\]
Cancelling the term 2 on the right side, we get
\[2\sin A\cos A=1\]
From the trigonometric table, $1=\sin {{90}^{0}}$
$2\sin A\cos A=\sin {{90}^{0}}$
$2\sin A\cos A=\sin \left( 2\times {{45}^{0}} \right)$
Now put $A={{45}^{0}}$, we get
$2\sin A\cos A=\sin \left( 2A \right)$
Hence the given expression is verified
Note: Doubling the $\sin x$ will not give you the value of $\sin 2x$ . Nor will taking half of $\sin x$, give you $\sin \left( \dfrac{x}{2} \right)$ . We can develop the double angle formulas directly by using the addition formulas for sine, cosine and tangent.
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