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TrigonometryQuestion and Answers: Page 2

Question Number 213887 Answers: 1 Comments: 0

Find amplitude, period, maximum and minimum value for function f(x)= 6 tan ((1/5)x)−8

$Find amplitude, period, maximum$ $and minimum value for function$ $f (x) = 6 \tan (\frac{1}{5} x) - 8$

Question Number 213861 Answers: 1 Comments: 0

Question Number 213741 Answers: 0 Comments: 1

(√(1−sin))

$\sqrt{1 - \sin}$

Question Number 213643 Answers: 3 Comments: 0

Find the maximum value of 3sin^2 x−8cosx+5=?

$F i n d t h e m a x i m u m v a l u e o f$ $3 {s i n}^{2} x - 8 c o s x + 5 = ?$

Question Number 213606 Answers: 2 Comments: 0

△ABC. 2a+b=2c. Find the minimum of (3/(sin C)) + (1/(tan A)).

$△ A B C . 2 a + b = 2 c . Find the minimum of$ $\frac{3}{\sin C} + \frac{1}{\tan A} .$

Question Number 213504 Answers: 1 Comments: 0

Question Number 213519 Answers: 2 Comments: 0

Question Number 213119 Answers: 2 Comments: 1

determiner a et b par ; AB ⊥BC { ((AM =5)),((AC =16)) :}

$determiner a et b par; AB ⊥ BC$ ${\begin{cases} AM = 5 \\ AC = 16 \end{cases}$

Question Number 213084 Answers: 4 Comments: 0

If 0 < x < (π/2) then prove sin(cosx) < cosx < cos(sinx). not using graph.

$If 0 < x < \frac{π}{2} then prove$ $\sin (\cos x) < \cos x < \cos (\sin x) .$ $not using graph .$

Question Number 213063 Answers: 0 Comments: 1

Area of [ACBD] ? {A ,B }∈ elipse verticale [AB]⊥MC M∈ elipse horizontale A ∈[elipse horizontale ∩elipse veeticale].

$Area of [ACBD] ?$ ${A, B} \in elipse verticale$ $[AB] ⊥ MC$ $M \in elipse horizontale$ $A \in [elipse horizontale \cap elipse veeticale] .$

Question Number 213054 Answers: 1 Comments: 1

Question Number 212777 Answers: 0 Comments: 0

question 212462 prove that ∫_0 ^∞ (dx/( (√(x^4 +25x^2 +160))))=∫_0 ^∞ (dx/( (√(x^4 −95x^2 +2560)))) my attempt (but is it a proof?): I_b ^a :=∫_0 ^∞ (dx/( (√(x^4 +ax^2 +b))))=(2/( π(b)^(1/4) agm (1, ((√(a+2(√b)))/(2(b)^(1/4) ))))) ∫_0 ^∞ (dx/( (√(x^4 +ax^2 +b))))= [t=2arctan (x/( (b)^(1/4) ))] =(1/( (b)^(1/4) ))∫_0 ^(π/2) (dt/( (√(1−((1/2)−(a/(4(√b))))sin^2 t))))= [K (k^2 ):=∫_0 ^(π/2) (dt/( (√(1−k^2 sin^2 t))))] =(1/( (b)^(1/4) ))K ((1/2)−(a/(4(√b)))) K (k^2 ) =(2/(πagm (1, (√(1−k^2 ))))) (1/( (b)^(1/4) ))K ((1/2)−(a/(4(√b)))) =(2/( π(b)^(1/4) agm (1, ((√(a+2(√b)))/(2(b)^(1/4) ))))) u_0 =p∧v_0 =q u_(n+1) =((u_n +v_n )/2)∧v_(n+1) =(√(u_n v_n )) agm (p, q) =lim_(n→∞) u_n =lim_(n→∞) v_n I_(160) ^(25) =(1/( π((10))^(1/4) agm (1, ((√(25+8(√(10))))/(4((10))^(1/4) ))))) I_(2560) ^(−95) =(1/(2π((10))^(1/4) agm (1, ((√(−95+32(√(10))))/(8((10))^(1/4) ))))) I_(160) ^(25) =I_(2560) ^(−95) ⇔ agm (1, ((√(25+8(√(10))))/(4((10))^(1/4) )))=2agm (1, ((√(−95+32(√(10))))/(8((10))^(1/4) ))) which is true

$question 212462$ $prove that$ $\underset{0}{\int^{\infty}} \frac{d x}{\sqrt{x^{4} + 25 x^{2} + 160}} = \underset{0}{\int^{\infty}} \frac{d x}{\sqrt{x^{4} - 95 x^{2} + 2560}}$ $my attempt (but is it a proof ?) :$ $I_{b}^{a} := \underset{0}{\int^{\infty}} \frac{d x}{\sqrt{x^{4} + {a x}^{2} + b}} = \frac{2}{π \sqrt[4]{b} agm (1, \frac{\sqrt{a + 2 \sqrt{b}}}{2 \sqrt[4]{b}})}$ $\underset{0}{\int^{\infty}} \frac{d x}{\sqrt{x^{4} + {a x}^{2} + b}} =$ $[t = 2 \arctan \frac{x}{\sqrt[4]{b}}]$ $= \frac{1}{\sqrt[4]{b}} \underset{0}{\int^{π / 2}} \frac{d t}{\sqrt{1 - (\frac{1}{2} - \frac{a}{4 \sqrt{b}}) \sin^{2} t}} =$ $[K (k^{2}) := \underset{0}{\int^{π / 2}} \frac{d t}{\sqrt{1 - k^{2} \sin^{2} t}}]$ $= \frac{1}{\sqrt[4]{b}} K (\frac{1}{2} - \frac{a}{4 \sqrt{b}})$ $K (k^{2}) = \frac{2}{π agm (1, \sqrt{1 - k^{2}})}$ $\frac{1}{\sqrt[4]{b}} K (\frac{1}{2} - \frac{a}{4 \sqrt{b}}) = \frac{2}{π \sqrt[4]{b} agm (1, \frac{\sqrt{a + 2 \sqrt{b}}}{2 \sqrt[4]{b}})}$ $u_{0} = p \land v_{0} = q$ $u_{n + 1} = \frac{u_{n} + v_{n}}{2} \land v_{n + 1} = \sqrt{u_{n} v_{n}}$ $agm (p, q) = \lim_{n \to \infty} u_{n} = \lim_{n \to \infty} v_{n}$ $I_{160}^{25} = \frac{1}{π \sqrt[4]{10} agm (1, \frac{\sqrt{25 + 8 \sqrt{10}}}{4 \sqrt[4]{10}})}$ $I_{2560}^{- 95} = \frac{1}{2 π \sqrt[4]{10} agm (1, \frac{\sqrt{- 95 + 32 \sqrt{10}}}{8 \sqrt[4]{10}})}$ $I_{160}^{25} = I_{2560}^{- 95}$ $\Leftrightarrow$ $agm (1, \frac{\sqrt{25 + 8 \sqrt{10}}}{4 \sqrt[4]{10}}) = 2 agm (1, \frac{\sqrt{- 95 + 32 \sqrt{10}}}{8 \sqrt[4]{10}})$ $which is true$