3-sin-cos-4-6-sin-cos-2-4-sin-6-cos-6-

Question Number 207516 by MATHEMATICSAM last updated on 17/May/24

3 {(\sin θ - \cos θ)}^{4} + 6 {(\sin θ + \cos θ)}^{2}

+ 4 (\sin^{6} θ + \cos^{6} θ) = ?

Commented by A5T last updated on 17/May/24

D i d y o u e d i t t h i s q u e s t i o n t o c h a n g e s o m e t h i n g ?

Answered by A5T last updated on 17/May/24

3 {[{s i n}^{2} θ + {c o s}^{2} θ - 2 s i n θ c o s θ]}^{2} + 6 (1 + 2 s i n θ c o s θ)

+ 4 [{({s i n}^{2} θ + {c o s}^{2} θ)}^{3} - 3 {s i n}^{2} θ {c o s}^{2} θ ({s i n}^{2} θ + {c o s}^{2} θ)]

= 3 {[1 - 2 s i n θ c o s θ]}^{2} + 6 + 6 s i n 2 θ + 4 [1 - 3 {s i n}^{2} θ {c o s}^{2} θ]

= 3 [1 + 4 {s i n}^{2} θ {c o s}^{2} θ - 4 s i n θ c o s θ] + 6 + 4

- 12 {s i n}^{2} θ {c o s}^{2} θ + 6 s i n 2 θ

= 13 - 12 s i n θ c o s θ + 6 s i n 2 θ = 13

Commented by Frix last updated on 17/May/24

6 {(\sin θ + \cos θ)}^{2} \neq 6 {(1)}^{2}

Commented by A5T last updated on 17/May/24

6 {({s i n}^{2} θ + {c o s}^{2} θ)}^{2} w a s t h e r e i n i t i a l l y, I s u p p o s e .

Commented by Frix last updated on 17/May/24

I {don}^{'} t know, I saw it like it is .

But I believe you, this also happened to me

before \dots

Answered by Frix last updated on 17/May/24

Let t = \tan θ

\dots = \frac{3 {(t - 1)}^{4}}{{(t^{2} + 1)}^{2}} + \frac{6 {(t + 1)}^{2}}{t^{2} + 1} + \frac{4 (t^{4} - t^{2} + 1)}{{(t^{2} + 1)}^{2}} = 13

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