sin cube theta plus cos cube theta by sin theta + cos theta=. (with clear explanation) ( very fast) I will mark you as brainliest
Answers
Answer:
sin3θ
=sin(2θ+θ)
=sin2θcosθ+cos2θsinθ
=(2sinθcosθ)cosθ+(1−2sin
2
θ)sinθ
=2sinθcos
2
θ+sinθ−2sin
3
θ
=2sinθ(1−sin
2
θ)+sinθ−2sin
3
θ
=2sinθ−2sin
3
θ+sinθ−2sin
3
θ
=3sinθ−4sin
3
θ.
From these formulas, we also have the following identities for \sin^3(\theta)sin
3
(θ) and \cos^3 (\theta)cos
3
(θ) in terms of lower powers:
\sin^3(\theta) = \frac{3 \sin (\theta) - \sin \left( 3 \theta \right) }{4},\quad \cos^3 (\theta) = \frac{\cos(3\theta) + 3 \cos (\theta)}{4}.
sin
3
(θ)=
4
3sin(θ)−sin(3θ)
,cos
3
(θ)=
4
cos(3θ)+3cos(θ)
.
To remember the cosine formula, the trick that I like to use is to read cosine as "dollar." Then, we say
"Dollar thirty is equal to four dollar thirty minus three dollar."
\begin{array} {l l l l l } \$1. 30 & = \$ 4.30 & - \$ 3 \\ 1 \cos 3 \theta & = 4 \cos ^3 \theta & - 3 \cos \theta \\ \end{array}
$1.30
1cos3θ
=$4.30
=4cos
3
θ
−$3
−3cosθ
Just leaving a mark here,
\begin{aligned} \tan x \tan(60^\circ - x) \tan (60^\circ + x) &=& \tan(3x) \\ 4\sin x \sin(60^\circ - x) \sin (60^\circ + x) &=&\sin(3x) \\ 4\cos x \cos(60^\circ - x) \cos(60^\circ + x) &=& \cos(3x) \\ \tan(3x) &=& \frac {3\tan x - \tan^3 x}{1 - 3\tan^2 x}. \end{aligned}
tanxtan(60
∘
−x)tan(60
∘
+x)
4sinxsin(60
∘
−x)sin(60
∘
+x)
4cosxcos(60
∘
−x)cos(60
∘
+x)
tan(3x)
=
=
=
=
tan(3x)
sin(3x)
cos(3x)
1−3tan
2
x
3tanx−tan
3
x
Prove that
\tan (6^\circ) = \tan (12^\circ) \tan(24^\circ) \tan(48^)