Asked by reduction formula trigonometry
sin130.tan60 ÷cos540.tan230.sin400
Answers
Answered by
Reiny
sin130° = sin 50°
tan60° = √3
cos540° = cos180 = -1
tan230° = tan50
sin400= sin40
sin130.tan60 ÷cos540.tan230.sin400
= (sin50°)(√3)/(-1) (tan50)(sin40)
= -√3sin50 (sin50/cos50)(sin40)
= -√3 sin40° sin^2 50° / cos50°
I see nothing special from here.
Were there brackets of sorts ?
tan60° = √3
cos540° = cos180 = -1
tan230° = tan50
sin400= sin40
sin130.tan60 ÷cos540.tan230.sin400
= (sin50°)(√3)/(-1) (tan50)(sin40)
= -√3sin50 (sin50/cos50)(sin40)
= -√3 sin40° sin^2 50° / cos50°
I see nothing special from here.
Were there brackets of sorts ?
Answered by
Steve
I think it was
sin130.tan60 ÷(cos540.tan230.sin400)
= (sin50 √3)/(-1 * sin50/cos50 * sin40)
= -√3 cos50/sin40
But since cos50 = sin40, the result is just
-√3
sin130.tan60 ÷(cos540.tan230.sin400)
= (sin50 √3)/(-1 * sin50/cos50 * sin40)
= -√3 cos50/sin40
But since cos50 = sin40, the result is just
-√3
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