show that (a) cos[2cos^(−1) (x) +sin^(−1) (x)]= −(√(1−x^2 )) (b) ((sinα + sinβ)/(cosα−cosβ))=cot(((β−α)/2)) (c) 2cos((π/3)+p)≊ 1−(√3) if p is small enough to neglect p^2 . (d) if θ =(1/2)sin^(−1) ((3/4)), show that sinθ−cosθ = ±(1/2) (e)write tan3A in terms of tanA (f) Factorise cosθ − cos3θ−cos5θ+cos7θ (g)i) verify that f(x)=((sin2θ+sin10θ)/(cos2θ+cos10θ))=((2tan3θ)/(1−tan^2 3θ)) ii) hence find in radians the general solution of f(x)=1 sir Forkum Michael


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 63426 by Rio Michael last updated on 04/Jul/19

$s h o w t h a t$ $(a) c o s [2 {c o s}^{- 1} (x) + {s i n}^{- 1} (x)] = - \sqrt{1 - x^{2}}$ $(b) \frac{s i n α + s i n β}{c o s α - c o s β} = c o t (\frac{β - α}{2})$ $(c) 2 c o s (\frac{π}{3} + p) ≊ 1 - \sqrt{3} i f p i s s m a l l e n o u g h t o n e g l e c t p^{2} .$ $(d) i f θ = \frac{1}{2} {s i n}^{- 1} (\frac{3}{4}), s h o w t h a t s i n θ - c o s θ = \pm \frac{1}{2}$ $(e) w r i t e t a n 3 A i n t e r m s o f t a n A$ $(f) F a c t o r i s e c o s θ - c o s 3 θ - c o s 5 θ + c o s 7 θ$ $(g) i) v e r i f y t h a t f (x) = \frac{s i n 2 θ + s i n 10 θ}{c o s 2 θ + c o s 10 θ} = \frac{2 t a n 3 θ}{1 - {t a n}^{2} 3 θ}$ $i i) h e n c e f i n d i n r a d i a n s t h e g e n e r a l s o l u t i o n o f f (x) = 1$ $s i r F o r k u m M i c h a e l$
Commented by Tony Lin last updated on 04/Jul/19

$(a)$ $l e t θ = {c o s}^{- 1} (x) \Rightarrow c o s θ = x$ $l e t \frac{π}{2} - θ = {s i n}^{- 1} (x) \Rightarrow s i n θ = \sqrt{1 - x^{2}}$ $c o s [2 θ + (\frac{π}{2} - θ)]$ $= c o s (\frac{π}{2} + θ)$ $= - s i n θ$ $= - \sqrt{1 - x^{2}}$
Commented by Tony Lin last updated on 04/Jul/19

$(b)$ $\frac{s i n α + s i n β}{c o s α - c o s β}$ $= \frac{2 s i n (\frac{α + β}{2}) c o s (\frac{α - β}{2})}{- 2 s i n (\frac{α + β}{2}) s i n (\frac{α - β}{2})}$ $= \frac{- c o s (\frac{α - β}{2})}{s i n (\frac{α - β}{2})}$ $= c o t (\frac{β - α}{2})$
Commented by Tony Lin last updated on 04/Jul/19

$(d)$ $θ = \frac{1}{2} {s i n}^{- 1} (\frac{3}{4})$ $2 θ = {s i n}^{- 1} (\frac{3}{4})$ $\Rightarrow \frac{3}{4} = s i n 2 θ$ $\Rightarrow \frac{3}{4} = 2 s i n θ c o s θ$ ${(s i n θ - c o s θ)}^{2}$ $= 1 - 2 s i n θ c o s θ$ $= \frac{1}{4}$ $\Rightarrow s i n θ - c o s θ = \pm \frac{1}{2}$
Commented by Tony Lin last updated on 04/Jul/19

$(e)$ $t a n 3 A$ $= t a n (A + 2 A)$ $= \frac{t a n A + t a n 2 A}{1 - t a n A t a n 2 A}$ $= \frac{t a n A + \frac{2 t a n A}{1 - {t a n}^{2} A}}{1 - \frac{2 {t a n}^{2} A}{1 - {t a n}^{2} A}}$ $= \frac{3 t a n A - {t a n}^{3} A}{1 - 3 {t a n}^{2} A}$
Commented by Rio Michael last updated on 05/Jul/19

$p l e a s e e x p l a i n t h e o c c u r a n c e o f$ ${(s i n θ - c o s θ)}^{2}$
	Answered by som(math1967) last updated on 04/Jul/19

	$g (i) \frac{s i n 2 θ + s i n 10 θ}{c o s 2 θ + c o s 10 θ} = \frac{2 s i n 6 θ c o s 4 θ}{2 c o s 6 θ c o s 4 θ} = t a n 6 θ$ $a g a i n t a n 6 θ = t a n 2.3 θ = \frac{2 t a n 3 θ}{1 - {t a n}^{2} 3 θ}$ $i i) A g a i n \frac{2 t a n 3 θ}{1 - {t a n}^{2} 3 θ} = 1$ $\Rightarrow {t a n}^{2} 3 θ + 2 t a n 3 θ = 1$ $\Rightarrow {t a n}^{2} 3 θ + 2 t a n 3 θ + 1 = 2$ $\Rightarrow {(t a n 3 θ + 1)}^{2} = 2$ $\Rightarrow t a n 3 θ = \sqrt{2} - 1 = t a n \frac{π}{8}$ $∴ 3 θ = n π + \frac{π}{8} ∴ θ = \frac{1}{3} (n π + \frac{π}{8})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum