I need help on these problems they are from my study guide and i don't have a clue on how to do these 3 problems and if anyone can show the steps and somewhat explain the steps so i can do it myself next time.
1.) sin25�‹cos35�‹sin�‹35
The answer is �ã3/2
2.) tan10�‹+tan20�‹/1-tan10�‹tan20�‹
The answer is �ã3/3
3.) cos(7ƒÎ/12)+cos(5ƒÎ/12)+sin7ƒÎ/12)sin(5ƒÎ12)
The answer is �ã3/2
2 answers
I don't know why the format is like this but the symbol behind the sin25 are degrees,the letter a on the answers are root, and the fl letter on number 3 are suppose to be PIE
1. First of all sin25cos35sin35 is NOT equal to √3/2
I know sin 60° = √3/2, so I looked at your expression to see what you might have meant.
since 25+35 = 60, you probably meant
sin25°cos35° + cos35°sin25°
which would be sin(25+35)° = sin60° = √3/2
2. again a typo, you meant
(tan10+tan20)/(1-tan10tan20))
by the tan(A+B) expansion
(tan10+tan20)/(1-tan10tan20))
= tan(10+20)°
=tan 30°
= 1/√3 , which is the same as your √3/3
3. once again a typo
Should have been
cos(7π/12)cos(5π/12) + sin(7π/12)sins5π/12)
= cos(7π/12 - 5π/12) , by the cos(A-B) expansion
= cos π/6 or cos 30°
=√3/2
I know sin 60° = √3/2, so I looked at your expression to see what you might have meant.
since 25+35 = 60, you probably meant
sin25°cos35° + cos35°sin25°
which would be sin(25+35)° = sin60° = √3/2
2. again a typo, you meant
(tan10+tan20)/(1-tan10tan20))
by the tan(A+B) expansion
(tan10+tan20)/(1-tan10tan20))
= tan(10+20)°
=tan 30°
= 1/√3 , which is the same as your √3/3
3. once again a typo
Should have been
cos(7π/12)cos(5π/12) + sin(7π/12)sins5π/12)
= cos(7π/12 - 5π/12) , by the cos(A-B) expansion
= cos π/6 or cos 30°
=√3/2
1 answer