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Hint- Here, we will be using formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ and identity ${\left( {\sin x} \right)^2} + {\left( {\cos x} \right)^2} = 1$.

To prove: \[{\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} = 2{\left( {\sin {\text{A}}} \right)^2} - 1 = 1 - 2{\left( {\cos {\text{A}}} \right)^2}\]

Let us simply the left most side of the above equation, we get

\[{\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left[ {{{\left( {\sin {\text{A}}} \right)}^2}} \right]^2} - {\left[ {{{\left( {\cos {\text{A}}} \right)}^2}} \right]^2}\]

Using formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, we get

\[ \Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left[ {{{\left( {\sin {\text{A}}} \right)}^2}} \right]^2} - {\left[ {{{\left( {\cos {\text{A}}} \right)}^2}} \right]^2} = \left[ {{{\left( {\sin {\text{A}}} \right)}^2} - {{\left( {\cos {\text{A}}} \right)}^2}} \right]\left[ {{{\left( {\sin {\text{A}}} \right)}^2} + {{\left( {\cos {\text{A}}} \right)}^2}} \right]\]

Now applying identity ${\left( {\sin x} \right)^2} + {\left( {\cos x} \right)^2} = 1$, we get

\[

\Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = \left[ {{{\left( {\sin {\text{A}}} \right)}^2} - {{\left( {\cos {\text{A}}} \right)}^2}} \right]\left[ 1 \right] = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} \\

\Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2}{\text{ }} \to {\text{(2)}} \\

\]

Now let us represent the complete RHS of equation (2) in terms of sine trigonometric function alone.

This can be done by using the identity ${\left( {\sin x} \right)^2} + {\left( {\cos x} \right)^2} = 1 \Rightarrow {\left( {\cos x} \right)^2} = 1 - {\left( {\sin x} \right)^2}$

\[

\Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} = {\left( {\sin {\text{A}}} \right)^2} - \left[ {1 - {{\left( {\sin {\text{A}}} \right)}^2}} \right] = {\left( {\sin {\text{A}}} \right)^2} - 1 + {\left( {\sin {\text{A}}} \right)^2} \\

\Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} = 2{\left( {\sin {\text{A}}} \right)^2} - 1{\text{ }} \to {\text{(3)}} \\

\]

Now let us represent the complete RHS of equation (2) in terms of cosine trigonometric function alone.

This can be done by using the identity ${\left( {\sin x} \right)^2} + {\left( {\cos x} \right)^2} = 1 \Rightarrow {\left( {\sin x} \right)^2} = 1 - {\left( {\cos x} \right)^2}$

\[

\Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} = \left[ {1 - {{\left( {\cos {\text{A}}} \right)}^2}} \right] - {\left( {\cos {\text{A}}} \right)^2} = 1 - {\left( {\cos {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} \\

\Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} = 1 - 2{\left( {\cos {\text{A}}} \right)^2}{\text{ }} \to {\text{(4)}} \\

\]

Now combining equations (3) and (4), we can write

\[{\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} = 2{\left( {\sin {\text{A}}} \right)^2} - 1 = 1 - 2{\left( {\cos {\text{A}}} \right)^2}\]

The above equation is the required equation which needs to be proved.

Note- In these types of problems, we simplify the one side of the equation which needs to be proved keeping in mind the trigonometric functions in which the final terms need to be represented.

To prove: \[{\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} = 2{\left( {\sin {\text{A}}} \right)^2} - 1 = 1 - 2{\left( {\cos {\text{A}}} \right)^2}\]

Let us simply the left most side of the above equation, we get

\[{\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left[ {{{\left( {\sin {\text{A}}} \right)}^2}} \right]^2} - {\left[ {{{\left( {\cos {\text{A}}} \right)}^2}} \right]^2}\]

Using formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, we get

\[ \Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left[ {{{\left( {\sin {\text{A}}} \right)}^2}} \right]^2} - {\left[ {{{\left( {\cos {\text{A}}} \right)}^2}} \right]^2} = \left[ {{{\left( {\sin {\text{A}}} \right)}^2} - {{\left( {\cos {\text{A}}} \right)}^2}} \right]\left[ {{{\left( {\sin {\text{A}}} \right)}^2} + {{\left( {\cos {\text{A}}} \right)}^2}} \right]\]

Now applying identity ${\left( {\sin x} \right)^2} + {\left( {\cos x} \right)^2} = 1$, we get

\[

\Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = \left[ {{{\left( {\sin {\text{A}}} \right)}^2} - {{\left( {\cos {\text{A}}} \right)}^2}} \right]\left[ 1 \right] = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} \\

\Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2}{\text{ }} \to {\text{(2)}} \\

\]

Now let us represent the complete RHS of equation (2) in terms of sine trigonometric function alone.

This can be done by using the identity ${\left( {\sin x} \right)^2} + {\left( {\cos x} \right)^2} = 1 \Rightarrow {\left( {\cos x} \right)^2} = 1 - {\left( {\sin x} \right)^2}$

\[

\Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} = {\left( {\sin {\text{A}}} \right)^2} - \left[ {1 - {{\left( {\sin {\text{A}}} \right)}^2}} \right] = {\left( {\sin {\text{A}}} \right)^2} - 1 + {\left( {\sin {\text{A}}} \right)^2} \\

\Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} = 2{\left( {\sin {\text{A}}} \right)^2} - 1{\text{ }} \to {\text{(3)}} \\

\]

Now let us represent the complete RHS of equation (2) in terms of cosine trigonometric function alone.

This can be done by using the identity ${\left( {\sin x} \right)^2} + {\left( {\cos x} \right)^2} = 1 \Rightarrow {\left( {\sin x} \right)^2} = 1 - {\left( {\cos x} \right)^2}$

\[

\Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} = \left[ {1 - {{\left( {\cos {\text{A}}} \right)}^2}} \right] - {\left( {\cos {\text{A}}} \right)^2} = 1 - {\left( {\cos {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} \\

\Rightarrow {\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} = 1 - 2{\left( {\cos {\text{A}}} \right)^2}{\text{ }} \to {\text{(4)}} \\

\]

Now combining equations (3) and (4), we can write

\[{\left( {\sin {\text{A}}} \right)^4} - {\left( {\cos {\text{A}}} \right)^4} = {\left( {\sin {\text{A}}} \right)^2} - {\left( {\cos {\text{A}}} \right)^2} = 2{\left( {\sin {\text{A}}} \right)^2} - 1 = 1 - 2{\left( {\cos {\text{A}}} \right)^2}\]

The above equation is the required equation which needs to be proved.

Note- In these types of problems, we simplify the one side of the equation which needs to be proved keeping in mind the trigonometric functions in which the final terms need to be represented.

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