We're all done and now the flood is gone! Finally!
Ham and Japeth won’t help me with my trig problem!
What are you guys in here getting loud about!?
Because my problem is harder than yours and Japeth’s.
No way!! You couldn’t even begin to solve my problem! It doesn’t even have numbers in it.
Alright, alright boys. Shem, we will start with you. What does your problem say?
My problem says: sinθ, given that cscθ = -5/9 … What does that even mean?
It’s simple! First, notice that sine (sin) and cosecant (csc) are reciprocals of each other. This means they're inverses of each other and their product is equal to 1. So, substitute -5/9 into the cscθ of the equation sinθ= 1/cscθ. Your equation will look like sinθ= 1/ (-5/9) and you'll solve this by using Keep Change Flip, where you keep the 1, change the division to multiplication, and flip your faction. Therefore cscθ = -5/9, then sinθ =-9/5. Simple as that! Your turn, Ham!
My problem says: Determine the sign of the trigonometric functions of an angle in standard position with the given measure 283°.
Alright, so according to All Students Take Calculus, since angle 283° is in quadrant IV, it falls under “Calculus,” or cosine and its reverse, secant. This means that 283° is positive in the quadrant I (All), cosine and secant and is negative in sine, cosecant, tangent, and cotangent. That’s it! Japeth, it’s your turn now.
Actually, mine is kind of similar to Ham’s. My problem is: tanθ > 0, cosθ < 0 .
To solve this problem, it is quite like Ham’s problem. tanθ > 0, cosθ < 0 reads as “tangent of theta is greater than 0 and cosine of theta is less than 0.” Greater than 0 means positive and less than 0 means negative. So with this information, find the quadrant that tangent can be positive and cosine can be negative at the same time. This will be quadrant III