
Show that \[{{\cos }^{2}}A-{{\sin }^{2}}A=2{{\cos }^{2}}A-1\].
Answer
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Hint: In this problem, we have to prove the given trigonometric expression. Here we can first take the left-hand side part and simplify and solve it to get the right-hand side part. We can replace the sine part in the left-hand side as \[{{\sin }^{2}}A=1-{{\cos }^{2}}A\], we can then simplify it by multiplying the signs and adding the similar terms to get the right-hand side part.
Complete step-by-step solution:
Here we have to prove the given trigonometric expression.
The given trigonometric expression is \[{{\cos }^{2}}A-{{\sin }^{2}}A=2{{\cos }^{2}}A-1\].
We can first take the left-hand side part and simplify and solve it to get the right-hand side part.
LHS = \[{{\cos }^{2}}A-{{\sin }^{2}}A\]
We can now take the sine part as we know that \[{{\sin }^{2}}A=1-{{\cos }^{2}}A\].
We can now replace the above formula for sine part in the left-hand side, we get
LHS = \[{{\cos }^{2}}A-\left( 1-{{\cos }^{2}}A \right)\]
Here, we can see that we have similar terms, we can first multiply the signs inside and outside the bracket, we get
LHS = \[{{\cos }^{2}}A-1+{{\cos }^{2}}A\]
We can see that, we have similar terms with similar sign, so we can add them, we get
LHS = \[2{{\cos }^{2}}A-1\]
We can see that,
LHS = RHS.
Therefore, \[{{\cos }^{2}}A-{{\sin }^{2}}A=2{{\cos }^{2}}A-1\]
Hence proved.
Note: We should also remember some of the trigonometric formulas and identities to be substituted and solved for the both sides. Here we have used the formula \[{{\sin }^{2}}A=1-{{\cos }^{2}}A\], as we have only cosine terms in the right-hand side. We have to choose the formula, in order to prove for the other side.
Complete step-by-step solution:
Here we have to prove the given trigonometric expression.
The given trigonometric expression is \[{{\cos }^{2}}A-{{\sin }^{2}}A=2{{\cos }^{2}}A-1\].
We can first take the left-hand side part and simplify and solve it to get the right-hand side part.
LHS = \[{{\cos }^{2}}A-{{\sin }^{2}}A\]
We can now take the sine part as we know that \[{{\sin }^{2}}A=1-{{\cos }^{2}}A\].
We can now replace the above formula for sine part in the left-hand side, we get
LHS = \[{{\cos }^{2}}A-\left( 1-{{\cos }^{2}}A \right)\]
Here, we can see that we have similar terms, we can first multiply the signs inside and outside the bracket, we get
LHS = \[{{\cos }^{2}}A-1+{{\cos }^{2}}A\]
We can see that, we have similar terms with similar sign, so we can add them, we get
LHS = \[2{{\cos }^{2}}A-1\]
We can see that,
LHS = RHS.
Therefore, \[{{\cos }^{2}}A-{{\sin }^{2}}A=2{{\cos }^{2}}A-1\]
Hence proved.
Note: We should also remember some of the trigonometric formulas and identities to be substituted and solved for the both sides. Here we have used the formula \[{{\sin }^{2}}A=1-{{\cos }^{2}}A\], as we have only cosine terms in the right-hand side. We have to choose the formula, in order to prove for the other side.
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