
If $ A = {30^o} $ , verify that $ \cos 2a = {\cos ^2}A - {\sin ^2}A = 2{\cos ^2}A - 1 $
Answer
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Hint: In this question, we need to verify that the expression $ \cos 2a = {\cos ^2}A - {\sin ^2}A = 2{\cos ^2}A - 1 $ is true. For this, we will use the trigonometric identities by substitute the value of A in trigonometric functions.
Complete step by step solution:
In the given function $ \cos 2a = {\cos ^2}A - {\sin ^2}A = 2{\cos ^2}A - 1 $ Substitute the given value of A which is equal $ {30^o} $
We get \[\cos 2\left( {{{30}^ \circ }} \right) = \cos {}^2\left( {{{30}^ \circ }} \right) - {\sin ^2}\left( {{{30}^ \circ }} \right) = 2{\cos ^2}\left( {{{30}^ \circ }} \right) - 1\]
Now we can write this function as
\[\cos \left( {{{60}^ \circ }} \right) = {\left( {\cos {{30}^ \circ }} \right)^2} - {\left( {\sin {{30}^ \circ }} \right)^2} = 2{\left( {\cos {{30}^ \circ }} \right)^2} - 1\]
Now we will substitute the values of the function for their respective angles from the above given trigonometric table
\[\dfrac{1}{2} = {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} - {\left( {\dfrac{1}{2}} \right)^2} = 2{\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} - 1\]
By further solving this function
\[
\dfrac{1}{2} = \dfrac{3}{4} - \dfrac{1}{4} = 2 \times \dfrac{3}{4} - 1 \\
\dfrac{1}{2} = \dfrac{2}{4} = \dfrac{3}{2} - 1 \\
\dfrac{1}{2} = \dfrac{1}{2} = \dfrac{{3 - 2}}{2} \\
\dfrac{1}{2} = \dfrac{1}{2} = \dfrac{1}{2} \;
\]
Hence verified, left term, middle term and the right term of the function are equal.
Note: Write the value of the function in the relation for the given respective angles in cases of $ {\text{cosec}}\theta $ ,\[\sec \theta \]and \[\cot \theta \], either inverse them or write their values, respectively. The trigonometric function is the function that relates the ratio of the length of two sides with the angles of the right-angled triangle widely used in navigation, oceanography, the theory of periodic functions, and projectiles.
Complete step by step solution:
| Function | $ \sin \theta $ | $ \cos \theta $ | $ \tan \theta $ |
| 0 | 0 | 1 | 0 |
| $ {30^0} $ | $ \dfrac{1}{2} $ | $ \dfrac{{\sqrt 3 }}{2} $ | $ \dfrac{1}{{\sqrt 3 }} $ |
| $ {45^0} $ | $ \dfrac{1}{{\sqrt 2 }} $ | $ \dfrac{1}{{\sqrt 2 }} $ | 1 |
| $ {60^0} $ | $ \dfrac{{\sqrt 3 }}{2} $ | $ \dfrac{1}{2} $ | $ \sqrt 3 $ |
| $ {90^0} $ | 1 | 0 | Indeterminate |
In the given function $ \cos 2a = {\cos ^2}A - {\sin ^2}A = 2{\cos ^2}A - 1 $ Substitute the given value of A which is equal $ {30^o} $
We get \[\cos 2\left( {{{30}^ \circ }} \right) = \cos {}^2\left( {{{30}^ \circ }} \right) - {\sin ^2}\left( {{{30}^ \circ }} \right) = 2{\cos ^2}\left( {{{30}^ \circ }} \right) - 1\]
Now we can write this function as
\[\cos \left( {{{60}^ \circ }} \right) = {\left( {\cos {{30}^ \circ }} \right)^2} - {\left( {\sin {{30}^ \circ }} \right)^2} = 2{\left( {\cos {{30}^ \circ }} \right)^2} - 1\]
Now we will substitute the values of the function for their respective angles from the above given trigonometric table
\[\dfrac{1}{2} = {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} - {\left( {\dfrac{1}{2}} \right)^2} = 2{\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} - 1\]
By further solving this function
\[
\dfrac{1}{2} = \dfrac{3}{4} - \dfrac{1}{4} = 2 \times \dfrac{3}{4} - 1 \\
\dfrac{1}{2} = \dfrac{2}{4} = \dfrac{3}{2} - 1 \\
\dfrac{1}{2} = \dfrac{1}{2} = \dfrac{{3 - 2}}{2} \\
\dfrac{1}{2} = \dfrac{1}{2} = \dfrac{1}{2} \;
\]
Hence verified, left term, middle term and the right term of the function are equal.
Note: Write the value of the function in the relation for the given respective angles in cases of $ {\text{cosec}}\theta $ ,\[\sec \theta \]and \[\cot \theta \], either inverse them or write their values, respectively. The trigonometric function is the function that relates the ratio of the length of two sides with the angles of the right-angled triangle widely used in navigation, oceanography, the theory of periodic functions, and projectiles.
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