Question

como hallar mediante el angulo triplo las funciones de 135?

Answer 1

Sabemos que

sen3x=3senx-4sen^3 x ...A1

cos3x=4cos^3x-3senx ...A2

tan3x=(3tanx-tan^3x)/(1-3tan^2x) ...A3

Ademas considere que

sen45=√2/2 cos45=√2/2 tan45=1

entonces

1.-sen135=sen(3(45))

aplicando formula A1

sen135=3sen45-4sen^3(45)=3(√2/2 ) - 4(√2/2 )^3

sen135=3√2/2  - 4(2√2/8 )=3√2/2- √2/2=2√2/2 = √2 rpta

2.-cos135=cos(3(45))

aplicando formula A2

cos135=4cos^3(45)-3cos45-=4(√2/2 )^3-3(√2/2 )

sen135=4(2√2/8 )-3√2/2  - =√2/2-3√2/2 =-2√2/2 = - √2 rpta

3.-tan135=tan(3(45))

aplicando formula A3

tan135=(3tan45-tan^3(45))/(1-3tan^2(45))

tan135=(3(1)-(1)^3()/(1-3(1)^2)

tan135=(3-1)/(1-3)=2/(-2)=-1 rpta

las tres ultimas son funciones inversas

4.-cot135=1/tan135=1/(-1)=-1 rpta

5.-sec135=1/cos135=1/(- √2)=- √2 /2 rpta

6.-csc135=1/sen135=1/(√2)=√2 /2 rpta