What is expand cos(8(t-pi/4))h(t-pi/4) ?

The solution to expand cos(8(t-pi/4))h(t-pi/4) is thcos(8t-2pi)-(pihcos(8t-2pi))/4

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expand cos(8(t-pi/4))h(t-pi/4)

expand

cos (8 (t - \frac{π}{4})) h (t - \frac{π}{4})

t h cos (8 t - 2 π) - \frac{πh cos ( 8 t - 2 π )}{4}

Solution steps

cos (8 (t - \frac{π}{4})) h (t - \frac{π}{4})

= h cos (8 (t - \frac{π}{4})) (t - \frac{π}{4})

Apply the distributive law:

a (b - c) = ab - a c

a = cos (8 (t - \frac{π}{4})) h, b = t, c = \frac{π}{4}

= cos (8 (t - \frac{π}{4})) h t - cos (8 (t - \frac{π}{4})) h \frac{π}{4}

= t h cos (8 (t - \frac{π}{4})) - \frac{π}{4} h cos (8 (t - \frac{π}{4}))

Simplify

t h cos (8 (t - \frac{π}{4})) - \frac{π}{4} h cos (8 (t - \frac{π}{4})) : t h cos (8 t - 2 π) - \frac{πh cos ( 8 t - 2 π )}{4}

= t h cos (8 t - 2 π) - \frac{πh cos ( 8 t - 2 π )}{4}

\frac{1}{x ^{2}} + 4 \frac{1}{x} - 5

\frac{x ^{2}}{s ( x + 2 ) ( x + 3 )}

(8 x + 6) (8 x - 6)^{'}

\frac{( p + 2 ) 2 + p x ^{5}}{10}

e^{t} \cdot u (t - 8)

What is expand cos(8(t-pi/4))h(t-pi/4) ?
The solution to expand cos(8(t-pi/4))h(t-pi/4) is thcos(8t-2pi)-(pihcos(8t-2pi))/4