Question

Please someone help me. It's about proving related to transformation of trigonometric ratios. Thank you!!​

Answer 1

Answer:

see explanation

Step-by-step explanation:

Using sum to product identities

cos x - cos y = - 2sin($\frac{x+y}{2}$ )sin($\frac{x-y}{2}$ ) = 2sin($\frac{x+y}{2}$ )sin($\frac{y-x}{2}$ )

cos x + cos y = 2cos($\frac{x+y}{2}$ )cos($\frac{x-y}{2}$ )

Note that

sin10° = sin(90 - 10)° = cos80°

Thus

$\frac{2sin(\frac{10+80}{2})sin(\frac{80-10}{2}) }{2cos(\frac{10+80}{2})cos(\frac{80-10}{2}) }$ ← cancel 2 from numerator/ denominator

= $\frac{sin45}{cos45}$ × $\frac{sin35}{cos35}$

= tan45° × tan35° { tan45° = 1 ]

= 1 × tan35°

= tan35° ← as required

Answer 2

Answer:    see proof below

Step-by-step explanation:

* Use the following Co-function Identity: sin A = cos(90 - A)

* Use the following Sum to Product Identities:

 cos x - cos y = 2 sin [(x + y)/2] · sin [(x - y)/2]

 cos x + cos y = 2 cos [(x + y)/2] · cos [(x - y)/2]

* Use the Unit Circle to evaluate cos 45 = sin 45 = √2/2

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \qquad \qquad \quad \dfrac{\cos 10-\sin 10}{\cos 10+\sin 10}\\\\\\\text{Co-function Identity:}\qquad \dfrac{\cos 10 -\cos (90-10)}{\cos 10+\cos (90-10)}\\\\\\\text{Product to Sum Identity:}\quad \dfrac{2\sin (\frac{10+80}{2})\sin (\frac{80-10}{2})}{2\cos (\frac{10+80}{2})\cos (\frac{80-10}{2})}\\\\\\\text{Simplify:}\qquad \qquad \qquad \qquad \dfrac{2\sin (45)\sin (35)}{2 \cos (45)\cos (35)}$

$.\qquad \qquad \qquad \qquad \qquad = \dfrac{2(\frac{\sqrt2}{2}) \sin (35)}{2(\frac{\sqrt2}{2}) \cos(35)}\\\\\\.\qquad \qquad \qquad \qquad \qquad = \tan (35)$

LHS = RHS:   tan 35 = tan 35   $\checkmark$