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Question Number 161322 by kapoorshah last updated on 16/Dec/21

Commented by cortano last updated on 16/Dec/21

$\sin (30 ° + 10 ° + α) = b$ $\frac{1}{2} \cos (10 ° + α) + \frac{1}{2} \sqrt{3} \sin (10 ° + α) = b$ $\cos (10 ° + α) + \sqrt{3} \sin (10 ° + α) = 2 b$ $\sqrt{3} \sqrt{1 - \cos^{2} t} = 2 b - \cos t; t = 10 ° + α$ $3 - 3 \cos^{2} t = 4 b^{2} - 4 b \cos t + \cos^{2} t$ $4 \cos^{2} t - 4 b \cos t + 4 b^{2} - 3$ $\cos t = \frac{4 b \pm \sqrt{16 b^{2} - 16 (4 b^{2} - 3)}}{8}$ $\cos t = \frac{b \pm \sqrt{3 - 3 b^{2}}}{2}$
Commented by MJS_new last updated on 16/Dec/21

$the calculation is right but for any given$ $angle α the answer is \frac{b - \sqrt{3 - 3 b^{2}}}{2} xor \frac{b + \sqrt{3 - 3 b^{2}}}{2}$ $and thus you must clearify your answer$
Commented by cortano last updated on 16/Dec/21

$i t d e p e n d f o r v a l u e b$
Commented by MJS_new last updated on 17/Dec/21

$this is what I think was missing$ $let 0 ° ⩽ α < 360 °$ $0 ⩽ α < 50 ° \lor 230 ° ⩽ α < 360 ° \Rightarrow \cos (α + 10 °) = \frac{b + \sqrt{3 (1 - b^{2})}}{2}$ $50 ° ⩽ α < 230 ° \Rightarrow \cos (α + 10 °) = \frac{b - \sqrt{3 (1 - b^{2})}}{2}$
	Answered by 1549442205PVT last updated on 16/Dec/21
	$sin(40^0 +α)=sin[30^0 +(10^0 +α)] =sin30^0 cos(10^0 +α)+cos30^0 sin(10^0 +α) =(1/2)cos(10^0 +α)+((√3)/2)sin(10^0 +α)=b we have the system of equations: { ((x+(√3)y=2b)),((x^2 +y^2 =1)) :}with x=cos(10^0 +α),y=sin(10^0 +α) ⇒(2b−(√3)y)^2 +y^2 =1⇔4y^2 −4b(√3)y+4b^2 −1=0 △′=12b^2 −16b^2 +4=4−4b^2 .Hence y=((2b(√3)±2(√(1−b^2 )))/4).From that x=cos(10^0 +α)=b−(√3)y=2b−((6b±2(√(3−3b^2 )))/4) =((2b±2(√(3−3b^2 )))/4)=((b±(√(3−3b^2 )))/2)$
	$s i n (40^{0} + α) = s i n [30^{0} + (10^{0} + α)]$ $= s i n 30^{0} c o s (10^{0} + α) + c o s 30^{0} s i n (10^{0} + α)$ $= \frac{1}{2} c o s (10^{0} + α) + \frac{\sqrt{3}}{2} s i n (10^{0} + α) = b$ $w e h a v e t h e s y s t e m o f e q u a t i o n s :$ ${\begin{cases} x + \sqrt{3} y = 2 b \\ x^{2} + y^{2} = 1 \end{cases} w i t h x = c o s (10^{0} + α), y = s i n (10^{0} + α)$ $\Rightarrow {(2 b - \sqrt{3} y)}^{2} + y^{2} = 1 \Leftrightarrow 4 y^{2} - 4 b \sqrt{3} y + 4 b^{2} - 1 = 0$ $△^{'} = 12 b^{2} - 16 b^{2} + 4 = 4 - 4 b^{2} . H e n c e$ $y = \frac{2 b \sqrt{3} \pm 2 \sqrt{1 - b^{2}}}{4} . F r o m t h a t$ $x = c o s (10^{0} + α) = b - \sqrt{3} y = 2 b - \frac{6 b \pm 2 \sqrt{3 - 3 b^{2}}}{4}$ $= \frac{2 b \pm 2 \sqrt{3 - 3 b^{2}}}{4} = \frac{b \pm \sqrt{3 - 3 b^{2}}}{2}$
	Commented by 1549442205PVT last updated on 16/Dec/21

	$T h a n k y o u S i r, f o r s i n (40^{0} + α) e x i s t$ $b m u s t s a t i s f y ∣ b ∣⩽ 1$
	Commented by MJS_new last updated on 16/Dec/21

	$also here, read my comment above$
	Commented by mr W last updated on 16/Dec/21

	$y o u {d i d n}^{'} t u n d e r s t a n d M J S s i r .$ $s a y α = 3 °, \sin (40 ° + 3 °) = \sin 43 ° = b$ $\cos (10 ° + 3 °) = \cos 13 ° w h i c h c a n o n l y$ $h a v e o n e p a r t i c u l a r v a l u e, n o t t w o!$ $a s y o u g i v e w i t h \frac{b \pm \sqrt{3 - 3 b^{2}}}{2} . i . e .$ $o n l y + o r -, n o t \pm .$
	Commented by cortano last updated on 17/Dec/21

	$h o w i f α = 160 ° ?$ $\cos (α + 160 °) = \cos 170 ° < 0$
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