Question

If cos a = 3/5 & cos b = 5/13 then prove that cos square (a+b/2) = 16/65​

Answer 1

Answer:

given, cos α = \frac{3}{5}

5

3

and cos β = \frac{5}{13}

13

5

and α, β are acute angles.

cosα = 3/5 = b/h

so, b = 3 and h = 5

then, p = √(h² - b²) = √(5² - 3²) = 4

sinα = 4/5

cosβ = 5/13 = b/h

so, b = 5 and h = 13

then, p = √(13² - 5²) = 12

sinβ = 12/13

(a) sin^{2}(\frac{\alpha - \beta}{2}) = \frac{1}{65}sin

2

(

2

α−β

)=

65

1

we know , sin²x = (1 - cos2x)/2

so, sin^2\frac{(\alpha-\beta)}{2}=\frac{1-cos(\alpha-\beta)}{2}sin

2

2

(α−β)

=

2

1−cos(α−β)

= \frac{1}{2}[1-cos\alpha cos\beta - sin\alpha sin\beta]

2

1

[1−cosαcosβ−sinαsinβ]

= 1/2 [ 1 - 3/5 × 5/13 - 4/5 × 12/13 ]

= 1/2 [ 1 - 15/65 - 48/65 ]

= 1/2 [ 1 - 63/65 ]

= 1/65 = RHS

(b) cos^{2}(\frac{\alpha + \beta}{2}) = \frac{16}{65}cos

2

(

2

α+β

)=

65

16

we know , cos²x = (1 + cos2x)/2

now, LHS = cos^{2}\left(\frac{\alpha + \beta}{2}\right)=\frac{1+cos(\alpha+\beta)}{2}cos

2

(

2

α+β

)=

2

1+cos(α+β)

= \frac{1+cos\alpha cos\beta - sin\alpha sin\beta}{2}

2

1+cosαcosβ−sinαsinβ

= 1/2 × [ 1 + 3/5 × 5/13 - 4/5 × 12/13 ]

= 1/2 × [ 1 + 15/65 - 48/65 ]

= 1/2 × [ 1 - 33/65 ]

= 1/2 × [ 32/65 ]

= 16/65 = RHS

Step-by-step explanation:

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