Cos A + Cos B, an important cosine function identity in trigonometry, is offered to uncover the amount of values of cosine role for angles A and B. It is one of the sum to product formulas supplied to represent the sum of cosine duty for angles A and also B into their product form. The result for Cos A + Cos B is offered as 2 cos ½ (A + B) cos ½ (A - B).

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Let us know the Cos A + Cos B formula and also its evidence in detail using addressed examples.

1.What is Cos A + Cos B identification in Trigonometry?
2.Cos A + Cos B sum to Product Formula
3.Proof of Cos A + Cos B Formula
4.How to apply Cos A + Cos B Formula?
5.FAQs on Cos A + Cos B

The trigonometric identification Cos A + Cos B is supplied to represent the sum of the cosine of angle A and also B, Cos A + Cos B in the product type using the compound angles (A + B) and (A - B). We will research the Cos A + Cos B formula in detail in the adhering to sections.


The Cos A + Cos B sum to product formula in trigonometry for angle A and also B is offered as,

Cos A + Cos B = 2 cos ½ (A + B) cos ½ (A - B)

Here, A and also B are angles, and (A + B) and (A - B) space their compound angles.

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We can offer the proof of Cos A + Cos B trigonometric formula utilizing the growth of cos(A + B) and cos(A - B) formula. Together we declared in the previous section, we create Cos A + Cos B = 2 cos ½ (A + B) cos ½ (A - B).

Let united state assume that (α + β) = A and also (α - β) = B. We know, utilizing trigonometric identities,

2α = A + B⇒ α = (A + B)/2

2β = A - B⇒ β = (A - B)/2

½ = cos α cos β, for any angles α and also β.

= 2 cos α cos β

⇒ Cos A + Cos B = 2 cos ½(A + B) cos ½(A - B)

Hence, proved.


We can apply the Cos A + Cos B formula as a amount to the product identity to make the calculation much easier when it is an overwhelming to find the cosine of offered angles. Allow us know its application making use of the instance of cos 60º + cos 30º. We will settle the worth of the offered expression through 2 methods, making use of the formula and by directly applying the values, and also compare the results. Have a look at the below-given steps.

Compare the angle A and B v the offered expression, cos 60º + cos 30º. Here, A = 60º, B = 30º.Solving using the expansion of the formula Cos A + Cos B, provided as, Cos A + Cos B = 2 cos ½ (A + B) cos ½ (A - B), us get,Cos 60º + Cos 30º = 2 cos ½ (60º + 30º) cos ½ (60º - 30º) = 2 cos 45º cos 15º = 2 (1/√2) ((√3 + 1)/2√2) = (√3 + 1)/2.Also, we know that cos 60º + cos 30º = (1/2 + √3/2) = (1 + √3)/2.

Hence, the result is verified.

Related object on Cos A + Cos B:

Let us have actually a look at a couple of examples to recognize the concept of cos A + cos B better.


Example 2: discover the worth of cos 160º + cos 20º.

Solution:

We know, Cos A + Cos B = 2 cos ½ (A + B) cos ½ (A - B)

Here, A = 160º, B = 20º

cos 160º + cos 20º = 2 cos ½ (160º + 20º) cos ½ (160º - 20º)

= 2 cos 90º cos 70º

= 0 <∵ cos 90º = 0>


Example 3: utilizing cos A + cos B, prove that (sin A + sin B)(sin A - sin B) = - (cos A + cos B)(cos A - cos B).

Solution:

Let us rearrange the provided expression.

(sin A + sin B)(sin A - sin B) = - (cos A + cos B)(cos A - cos B) deserve to be created as, (sin A + sin B)/(cos A + cos B) = -(cos A - cos B)/(sin A - sin B)

Here, L.H.S. = (sin A + sin B)/(cos A + cos B)

= <2 sin ½ (A + B) cos ½ (A - B)>/<2 cos ½ (A + B) cos ½ (A - B)>

= sin ½ (A + B)/cos ½ (A + B)

R.H.S. = -(cos A - cos B)/(sin A - sin B)

-<- 2 sin ½ (A + B) sin ½ (A - B)>/<2 cos ½ (A + B) sin ½ (A - B)>

= -<- sin ½ (A + B)>/

= sin ½ (A + B)/cos ½ (A + B)

⇒ L.H.S. = R.H.S.

Hence, proved.


Example 4: Verify the given expression using development of Cos A + Cos B: cos 70º + sin 70º = √2 cos 25º

Solution:

We have, L.H.S. = cos 70º + sin 70º

Since, sin 70º = sin(90º - 20º) = cos 20º

⇒ cos 70º + sin 70º = ⇒ cos 70º + cos 20º

Using Cos A + Cos B = 2 cos ½ (A + B) cos ½ (A - B)

⇒ cos 70º + cos 20º = 2 cos ½ (70º + 20º) cos ½ (70º - 20º)

= 2 cos 45º cos 25º

= √2 cos 25º

= R.H.S.

Hence, verified.


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