Question

Question #bc254

Answer 1

The answers are `(1)=4cosa` and `tan15=2-sqrt3`

Explanation:

Hola!

`sin^2a+cos^2a=1`

`sin(2a)=2sinacosa`

`:.`

`sin(2a)/(1-cos^2a)*((sin2a)/cosa)`

`=(2sinacosa)^2/((sin^2a*cosa))`

`(4cancel(sin^2a)cos^2a)/(cancel(sin^2a)cosa)`

`=4cosa`

`"second part"`

`tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)`

Let `alpha=60^@`, `=>`, `tan60=sqrt3`

and `beta=45^@`, `=>`, `tan45=1`

Therefore,

`tan(60-45)=tan15=(tan60-tan45)/(1+tan60tan45)`

`=(sqrt3-1)/(1+sqrt3)`

`=(sqrt3-1)/(sqrt3+1)*(sqrt3-1)/(sqrt3-1)`

`=(sqrt3-1)^2/(3-1)`

`=(3+1-2sqrt3)/(2)`

`=(4-2sqrt3)/2`

`=2-sqrt3`

Espero que eso lo haya ayudado!

Answer 2

Por favor ver mas abajo.

Explanation:

Ejercicio 1 (nota: se dice sen "sin" en Socratic):

Formula de angulos dobles:

`\sin 2a = 2\sin a\cos a`

`color(blue)({\sin 2a}/{1-\cos^2 a}={2\sin a\cos a}/{\sin^2a}={2\cos a}/{\sin a})`

`color(red)({\sin 2a}/{\cos a}={2\sin a\cos a}/\cos a=2\sin a`

`color(blue)({2\cos a}/{\sin a})×color(red)(2\sin a)=4\cos a`

Ejercicio 2:

Formula de la tangente de diferencia:

`\tan(a-b)={\tan a-\tan b}/{1+\tan a\tan b}`

`\tan(15°)=\tan(45°-30°)={\tan 45°-\tan 30°}/{1+\tan 45°\tan 30°}`

`={1-{1}/{\sqrt{3}}}/{1+(1×{1}/{\sqrt{3}})}`

`={sqrt{3}-1}/{sqrt{3}+1}`

`={(sqrt{3}-1)×(sqrt{3}-1)}/{3-1}`

`={4-2\sqrt{3}}/2=2-\sqrt{3}`