psin-con-2-a-pcos-sin-2-b-p-0-0-pi-2-prove-p-a-2-b-2-3-2-ab-

Question Number 202295 by liuxinnan last updated on 24/Dec/23

p s i n θ {c o n}^{2} θ = a

p c o s θ {s i n}^{2} θ = b

p \neq 0 θ \in (0, \frac{π}{2})

p r o v e p = \frac{{(a^{2} + b^{2})}^{\frac{3}{2}}}{a b}

Answered by 1990mbodji last updated on 24/Dec/23

\underset{―}{E x e r c i c e} :

p s i n θ {c o s}^{2} θ = a e t p c o s θ {s i n}^{2} θ = b o \overset{`}{u} θ \in] 0; \frac{π}{2} [.

M o n t r o n s q u e p = \frac{{(a^{2} + b^{2})}^{\frac{3}{2}}}{a b} .

a b = p^{2} {s i n}^{3} θ {c o s}^{3} θ e t a^{2} + b^{2} = p^{2} {s i n}^{2} θ {c o s}^{2} θ ({c o s}^{2} θ + {s i n}^{2} θ)

\Rightarrow a b p = {(p s i n θ c o s θ)}^{3} e t a^{2} + b^{2} = {(p s i n θ c o s θ)}^{2}

\Rightarrow p s i n θ c o s θ = {(a b p)}^{\frac{1}{3}} e t p s i n θ c o s θ = {(a^{2} + b^{2})}^{\frac{1}{2}}

\Rightarrow a b p = {(a^{2} + b^{2})}^{\frac{3}{2}} \Rightarrow p = \frac{{(a^{2} + b^{2})}^{\frac{3}{2}}}{a b} .

Commented by MathematicalUser2357 last updated on 26/Dec/23

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