Sujet: trig identity liste de sujets
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34 articles trouvés pour ce sujet.

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	A trig identity		2015-10-28
		Gail pose la question : verify sec(x) =cot(x)+tan(x)/csc(x), the csc(x) is under both the whole term cot(x) + tan(x) Penny Nom lui répond.
	A trig identity		2013-01-04
		Tehmas pose la question : Prove sinC+sinD=2sin(C+D/2)cos(C-D/2) Harley Weston lui répond.
	A trig identity		2012-01-08
		Joe pose la question : Prove this Trig. Identity : ((cos2θ + sinθ-1) / tanθ ) + sin2θ = cosθ Penny Nom lui répond.
	Tan(x+pi)-tan(pi-x)=2tan(x).		2011-01-09
		Steven pose la question : Verify the identity: Tan(x+pi)-tan(pi-x)=2tan(x) Penny Nom lui répond.
	A trig identity		2010-01-05
		Christine pose la question : I need help with this whole paper here is one problem csc^2x-cot^2x/1-sin^2x=sec^2x by the ^2 i mean squared pleease Penny Nom lui répond.
	A trig identity		2009-07-10
		sam pose la question : I need to prove: tan x (cot x + tan x) = sec2 x (the "sec2" is "sec squared") I am totally stuck!! Thanks! Stephen La Rocque and Harley Weston lui répond.
	Sec(2x)		2008-11-23
		Evan pose la question : Prove that this is an identity... sec2x = sec^2x / 2 - sec^2x Harley Weston lui répond.
	A trig Identity		2008-11-13
		Rebecca pose la question : Question from Rebecca, a student: How do I prove:      1/sin@                csc@                -2sin@ ________      +       ______       =    _____ 1 + CSC@         1-1/sin@                 cos^@ Penny Nom lui répond.
	A trig identity		2008-11-03
		Student pose la question : verify cos(2x)=(cot(x)-tan(x))/(cot(x)+tan(x)) Penny Nom lui répond.
	TRIG IDENTITIES		2008-08-17
		VINCE pose la question : HOW DO YOU SIMPLIFY THE FOLLOWING:- 2sin3(theta)cos3(theta)? Regards Harley Weston lui répond.
	cos3A=-1/3 find cos6A		2008-05-15
		mike pose la question : cos3A=-1/3 find cos6A Penny Nom lui répond.
	(1-tanx) / (1+tanx) = (1-sin2x) / cos2x		2008-02-14
		isaiah pose la question : i'm having trouble proving 1-tanx / 1+tanx = 1-sin2x / cos2x, can anyone help me from going crazy? Penny Nom lui répond.
	A trig identity		2007-11-10
		James pose la question : I need to prove trig identities, and I can't figure this one out 1 + sinx - cosx         1 - cosx -------------------- = ------------- 1 + sinx + cosx         sinx Chris Fisher lui répond.
	A trig identity		2006-12-14
		Zamira pose la question : i need help with this proof: ((cot(x)-tan(x))/ cot(x) + tan(x)) = cos (x) Penny Nom lui répond.
	Sin 2x / 1-Cos 2x = 2 Csc 2x - Tan x		2006-12-06
		Mark pose la question : I need to prove the following: Sin 2x / 1-Cos 2x = 2 Csc 2x - Tan x Steve La Rocque lui répond.
	tan1/2x + cot1/2x = 2cscx		2006-11-04
		Jazmin pose la question : Prove tan1/2x + cot1/2x = 2cscx Penny Nom lui répond.
	A trig identity		2006-06-04
		Courtney pose la question : Show that this equation is an identity: (sin x + sin 2X)/(cos x - cos 2x) = cot(x/2) Stephen La Rocque lui répond.
	A trig identity		2006-03-28
		Iqbal pose la question : Could you please prove, [tan(x)-sin(x)]/sin3(x)=sec(x)/[1+cos(x)] Stephen La Rocque lui répond.
	A pair of trig equations		2005-12-05
		Kevin pose la question : I am trying to solve for A and B but haven't been able to find a trig identity that will help me. 1.414 = .5cosA + cosB .5 = .5sinA - sinB how do I solve this? Penny Nom lui répond.
	A trig identity		2005-03-06
		Patrick pose la question : im trying to verify this identity: sin(x)+sin(5x) -------------------- = tan(3x) cos(x)+cos(5x) Chris Fisher lui répond.
	A trig identity		2003-05-20
		Patty pose la question : Please help with the following 1/ tanx + cotx = sinxcosx Penny Nom lui répond.
	A trig identity		2003-02-22
		Ron pose la question : We have spent hours trying to solve the following identity without success. Can you give us some hints as to how it is done? 1 + tan(x) tan(2x) = tan(2x) cot(x) -1 Penny Nom lui répond.
	An identity?		2002-10-27
		Jenna pose la question : I can't prove 1-sinQ 1 ---- = tanQ + --- 1+sinQ cosQ Penny Nom lui répond.
	An identity(?)		2002-05-02
		A student pose la question : prove identity sin 2x/1-sinx= secx+1/secx Paul Betts lui répond.
	A trigonometric identity		2002-03-22
		Debby pose la question : I am stuck on a problem and wondering if you can help?? It is: Prove the following: sec2(X)+csc2(X) = sec2(X)csc2(X) Harley Weston lui répond.
	A trig identity		2001-07-27
		Jeff pose la question : prove this identity and show steps tan(x/2+pi/4)=secx+tanx Harley Weston lui répond.
	Trig identity crisis		2000-11-29
		Rhiannon pose la question : I have tried many times to find the answer to these problems but I can't I am in grade 12 tan(x)=csc2(x)-cot2(x) cos(x)/csc(x)-2sin(x)=tan(x)/1-tan(x) cos(x)[ tan2(x)1-1]/cos2(x)+sin2(x)=sec(x) Harley Weston lui répond.
	A trigonmetric identity		2000-05-15
		Caitlin pose la question : My name is Caitlin and my question is from 11th grade math and I'm a student my question is I need to solve this identity : cos x csc x - sin x sec x = 2 cot 2x Paul Betts lui répond.
	A trig identiry		2000-02-23
		Ashlee pose la question : I am having problems doing this problem. Can you help? Verify the identity: (1+cot2X)(1-cos2X)=2 Harley Weston lui répond.
	A trig identity		2000-02-17
		Eric pose la question : Question: How do I solve this problem? sin3x cos3x _____ - _____ = 2 sinx cosx Chris Fisher lui répond.
	Proving a trigonometric identitiy		1999-12-17
		Ryan pose la question : I need to figure out how to prove that sec2x + csc2x = sec2x csc2x. I am not sure where to start out with it and whether I should use reciprocal, quotient, or pythagorean. Penny Nom lui répond.
	A trig problem		1999-06-03
		Stu Barnes pose la question : cos(theta) / 1+ sin(theta)=sec(theta)-tan(theta) I've being having trouble with this one on my correspondance course. Harley Weston lui répond.
	A trig identity		1996-03-11
		Azmat Hussain pose la question : Is there an easier/another way to prove the trig identity cos(a+b) = cos(a)cos(b)-sin(a)sin(b)? Penny Nom lui répond.
	Trig identities		1995-11-30
		Azmat pose la question : Why do we work on the two sides of a trig identity separately? Harley Weston lui répond.

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