# Thread:

1. ## [SOLVED] Verify that the equation is an identity.

$Sin 4\alpha = 4 Sin\alpha Cos\alpha Cos 2\alpha$

2. Originally Posted by something3k
$Sin 4\alpha = 4 Sin\alpha Cos\alpha Cos 2\alpha$
start on the left hand side and use the double angle formula for sine: $\sin 2 \alpha = 2 \sin \alpha \cos \alpha$

note that $\sin4 \alpha = \sin [2(2 \alpha)]$

3. so would that equal $4 \sin\alpha \cos\alpha?$

4. Originally Posted by something3k
so would that equal $4 \sin\alpha \cos\alpha?$
no, follow the rule: sin(2A) = 2*sinA*cosA

5. Hello, something3k!

Identity: . $2\!\cdot\!\sin\theta\!\cdot\!\cos\theta \:=\:\sin2\theta$

$\sin4\alpha \:= \:4\!\cdot\!\sin\alpha\!\cdot\!\cos\alpha\!\cdot\! \cos2\alpha$
Or start on the right side . . .

$4\!\cdot\!\sin\alpha\!\cdot\!\cos\alpha\!\cdot\!\c os2\alpha \;=\;2\cdot\underbrace{2\!\cdot\!\sin\alpha\!\cdot \!\cos\alpha}_{\text{This is }\sin2\alpha}\cdot\cos2\alpha$

. . . . . . . . . . . . . $= \;\underbrace{2\!\cdot\!\sin2\alpha\!\cdot\!\cos2\ alpha}$

. . . . . . . . . . . . . $=\qquad \sin4\alpha$

6. ahh i seee, thanks a lot you guys i am thinking a little too hard.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8014259338378906, "perplexity_flag": "middle"}