Question

Зная, что tg t=a, найдите sin t cos t, sin^4 t

Answer 1

I hope this helps you

Answer 2

A) a=sin  t/cos t=(sint*cost)|cos^2t

sin t * cos t= a*cos^2 t

a^2=(1/cos^2 t)-1     cos^2t=1/(a^2+1)

sin t * cos t= a/(a^2+1)

B)  sin^4 t=(1-cos^ t)^2

      sin^4 t=(1-1/(a^2+1))^2

     sin^4 t = a^4/(a^2+1))^2

Answer 3

If we know that $\tan(t) = a$, we can use trigonometric identities to find $\sin(t)\cos(t)$ and $\sin^4(t)$.

1. Finding $\sin(t)\cos(t)$: We know that $\tan(t) = \frac{\sin(t)}{\cos(t)} = a$. Cross-multiplying, we get $\sin(t) = a\cos(t)$. Now, let's use the Pythagorean identity: $\sin^2(t) + \cos^2(t) = 1$. Replacing $\sin(t)$ with $a\cos(t)$, we get: $(a\cos(t))^2 + \cos^2(t) = 1$. Solving for $\cos(t)$: $a^2\cos^2(t) + \cos^2(t) = 1$. $\cos^2(t)(a^2 + 1) = 1$. $\cos^2(t) = \frac{1}{a^2 + 1}$. Now, we can find $\sin(t)$: $\sin(t) = a\cos(t) = a\sqrt{\frac{1}{a^2 + 1}}$.

Therefore, $\sin(t)\cos(t) = a\sqrt{\frac{1}{a^2 + 1}} \cdot \sqrt{\frac{1}{a^2 + 1}} = \frac{a}{\sqrt{a^2 + 1}}$.

2. Finding $\sin^4(t)$: We can use the double-angle formula for sine to express $\sin^4(t)$ in terms of $\cos(t)$. Recall that $\sin^2(t) = (1 - \cos^2(t))$, and we already found $\cos^2(t) = \frac{1}{a^2 + 1}$.

$\sin^2(t) = 1 - \frac{1}{a^2 + 1} = \frac{a^2}{a^2 + 1}$.

Now, squaring again: $\sin^4(t) = \left(\frac{a^2}{a^2 + 1}\right)^2 = \frac{a^4}{(a^2 + 1)^2}$.

So, $\sin^4(t) = \frac{a^4}{(a^2 + 1)^2}$.