Question

Considerando Seno x igual a 5/13, calcular Cosseno de (x + 60), Seno de (x - 30) e Tangente de (x + 45). por favor expliquem a respost​

Answer 1

São Relações Estabelecidas

cos(a+b)=cos(a)*cos(b)-sen(a)*sen(b)

cos(a-b)=cos(a)*cos(b)+sen(a)*sen(b)

sen(a-b)=sen(a)*cos(b)-sen(b)*cos(a)

sen(a+b)=sen(a)*cos(b)+sen(b)*cos(a)

tan(a+b)=sen(a+b)/cos(a+b)

sen²(x)+cos²(x)=1

30° ,45° e 60° são ângulos fundamentais , estes tem que decorar

sen(60°)=√3/2

cos(60°)=1/2

tan(60°)=√3

cos(30°)=√3/2

sen(30°)=1/2

tan(30°)=√3/3

sen(45°)=√2/2

cos(45°)=√2/2

tan(45°)=1

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Resolvendo :

sen(x)=5/13  ele é do 1ª ou 2ª quadrante

sen²(x)+cos²(x)=1

(5/13)²+cos²(x)=1

cos²(x)=1-25/169

cos²(x)=144/169 ==>cos(x)=±√(144/169)=±12/13

Se for do 1ª quadrante ==>sen(x)=5/13  e cos(x)=12/13

Se for do 2ª quadrante ==>sen(x)=5/13  e cos(x)=-12/13

Vou considerar o ângulo x do 1ª quadrante , esta informação teria que estar no texto.

Cosseno de (x + 60=?

cos(x+60)=cos(x)*cos(60)-sen(60)*sen(x)

cos(x+60)=12/13 *1/2-√3/2*5/13= 6/13-5√3/26

Seno de (x - 30)=?

sen(x-30)=sen(x)*cos(30)-cos(x)*sen(30)

sen(x-30)=5/13*√3/2-12/13 *1/2 =5√3/26

Tangente de (x + 45) =?

Tan  (x + 45)=[sen(x+45)]/cos(x+45]

###sen(x+45)=sen(x)*cos(45)+sen(45)*cos(x)

###sen(x+45)=5/13*√2/2+√2/2*12/13 =√2/2*(17/13)=17√2/26

###cos(x+45)=cos(45)*sen(45)-sen(x)*sen(x)

###cos(x+45)=√2/2*√2/2-5/13  *5/13

### =2/4-25/169=1/2-25/169=119/338

Tan  (x + 45)=[17√2/26]/[119/338]

Tan  (x + 45)=[17√2/26]* [338/119]

Tan  (x + 45) = 5746√2/3094

Answer 2

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Cosseno da soma

$\large\boxed{\boxed{\boxed{\boxed{\sf cos(a+b)=cos(a)\cdot cos(b)-sen(a)\cdot sen(b)}}}}$

Seno da diferença

$\large\boxed{\boxed{\boxed{\boxed{\sf sen(a-b)=sen(a)\cdot cos(b)-sen(b)\cdot cos(a)}}}}$

Tangente da soma

$\large\boxed{\boxed{\boxed{\boxed{\sf tg(a+b)=\dfrac{tg(a)+tg(b)}{1-tg(a)\cdot tg(b)}}}}}$

$\rm nota: vou~assumir~que~x~pertenc_{\!\!,}a~ao~1^{\underline o}~quadrante\\\sf sen(x)=\dfrac{5}{13}\\\sf sen^2(x)=\dfrac{25}{169}\\\sf cos^2(x)=1-sen^2(x)\\\sf cos^2(x)=\dfrac{169}{169}-\dfrac{25}{169}=\dfrac{144}{169}\\\sf cos(x)=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}\\\sf tg(x)=\dfrac{sen(x)}{cos(x)}\\\sf tg(x)=\dfrac{\frac{5}{\diagdown\!\!\!\!\!\!13}}{\frac{12}{\diagdown\!\!\!\!\!\!13}}=\dfrac{5}{12}$

$\sf cos(x+60^\circ)=cos(x)\cdot cos(60^\circ)-sen(x)\cdot sen(60^\circ)\\\sf cos(x+60^\circ)=\dfrac{12}{13}\cdot\dfrac{1}{2}-\dfrac{5}{13}\cdot\dfrac{\sqrt{3}}{2}\\\sf cos(x+60^\circ)=\dfrac{12}{26}-\dfrac{5\sqrt{3}}{26}\\\huge\boxed{\boxed{\boxed{\boxed{\sf cos(x+60^\circ)=\dfrac{12-5\sqrt{3}}{26}}}}}\blue{\checkmark}$

$\sf sen(x-30^\circ)=sen(x)\cdot cos(30^\circ)-sen(30^\circ)\cdot cos(x)\\\sf sen(x-30^\circ)=\dfrac{5}{13}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}\cdot\dfrac{12}{13}\\\sf sen(x-30^\circ)=\dfrac{5\sqrt{3}}{26}-\dfrac{12}{26}\\\huge\boxed{\boxed{\boxed{\boxed{\sf sen(x-30^\circ)=\dfrac{5\sqrt{3}-12}{26}}}}}\blue{\checkmark}$

$\sf tg(x+45^\circ)=\dfrac{tg(x)+tg(45^\circ)}{1-tg(x)\cdot tg(45^\circ)}\\\sf tg(x+45^\circ)=\dfrac{\frac{5}{12}+1}{1-\frac{5}{12}\cdot1}\\\sf tg(x+45^\circ)=\dfrac{\frac{5+12}{12}}{\frac{12-5}{12}}\\\sf tg(x+45^\circ)=\dfrac{\frac{17}{\diagdown\!\!\!\!\!\!12}}{\frac{7}{\diagdown\!\!\!\!\!\!12}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf tg(x+45^\circ)=\dfrac{17}{7}}}}}\blue{\checkmark}$