Show that \[{{\cos }^{2}}A-{{\sin }^{2}}A=2{{\cos }^{2}}A-1\].
Answer
589.2k+ views
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.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

